A Binomial Expression

Any algebraic expression consisting of only two terms is known as a binomial expression. It's
expansion in powers of x is known as the binomial expansion.
1
For example: (i) a + x (ii) a2 + (iii) 4x – 6y
x2
Binomial Theorem:
Such formula by which any power of a binomial expression can be expanded in the form of a series
is known as binomial theorem. When n is a positive integer,
(a + x)n = nC0an+nC1an–1x + nC2 an-2 x2 + … + nCr an–r xr + … + nCnxn,
where nC0, nC1, nC2 …. nCn are called Binomial co-efficients.

Proof: Let P(n) be the statement

P(n): (a + x)n = nC0an + nC1an – 1x + nC2an – 2 x2 + …. + nCn – 1axn – 1 + nCnxn.
For n = 1, we get (a + x)1 = 1C0a1 + 1C1a0x = a + x.
Hence P(1) is true. Let P (r) be true for some positive integers r.
That is (a + x)r = rC0ar + rC1ar – 1x + … + rCrxr.
For n = r + 1, P(r + 1): (a + x)r + 1 = (a + x)r (a + x)
= (rC0ar + rC1ar – 1x + rC2ar – 2x2 + …. + rCrxr) (a + x)
= rC0ar + 1 + (rC0 + rC1)arx + (rC1 + rC2)ar – 1x2 + …. + (rCr – 1 + rCr)axr + rCrxr + 1.
But rC0 = 1= r + 1C0, rCr = 1 = r + 1Cr + 1,
rC + rC = r + 1C , rC + rC r + 1C .
0 1 1 k k–1 = k

Hence P(r + 1): (a + x) r + 1 = C0a + r+1C1arx + r+1C2ar–1x2 + ...+ r+ 1Cr axr + r+1Cr + 1 xr + 1
r+1 r+1

Here P (r + 1) is true and by mathematical induction P (n) is true.

Writing n = 0, 1, 2, 3, 4, 5, we get the following known formulae:

(a + x)0 = 1
(a + x)1 = a + x
(a + x)2 = a2 + 2ax + x2
(a + x)3 = a3 + 3a2x + 3ax2 + x3
(a + x)4 = a4 + 4a3x + 6a2x2 + 4ax3 + x4
(a + x)5 = a5 + 5a4x + 10a3x2 + 10a2x3 + 5ax4 + x5.
We list the coefficients in these expansions as

1
1 1
1 2 1
1 3 3 1
1 4 5 4 1
1 5 10 10 5 1
These coefficients (binomial coefficients) in combinatorial form may be re−written as
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0C
0

1C 1C
0 1

2C 2C 2C
0 1 2

3C 3C 3C 3C
0 1 2 3

4C 4C 4C 4C 4C
0 1 2 3 4

5C 5C 5C 5C 5C 5C
0 1 2 3 4 5

so that (a + x)4 = 4C0a4 + 4C1a3x + 4C2a2x2 + 4C3ax3 + 4C4x4 etc.

7
 1
Illustration 1. Expand  x +  .
 x

7
 1
Solution: x + x 
 
1 1 1 1 1 1 1
= 7C0x7+7C1x6 x +7C2x5 2 +7C3x4 3 + 7C4x3 4 + 7C5x2 5 + 7C6 x 6 +7C7
x x x x x x7
35 21 7 1
= x7 + 7x5 + 21x3 + 35 x + + 3 + 5 + 7 .
x x x x

Properties of the Binomial Expansion

There are (n + 1) terms in the expansion of (a + b)n, the first and the last term being a n and bn
respectively. If nCx = nCy, then either x = y or x + y = n
n !
 nCr = nCn - r = . e.g. 12C9 = 12C3.
r ! (n − r ) !
The general term (which is (r + 1)th term) is = nCr an - r br. It is usually denoted by Tr+1.
In the summation notation, the binomial theorem is written as
n
( a + b )n = 
r =0
n
Cr an−r br .

n is called the index of the binomial. The values of the binomial coefficients are symmetrically
placed about the middle value if the index n is even, and about two middle values (which are equal)
if the index is odd. For example, if n = 4, the number of terms is 5 and binomial coefficients are 1,
4, 6, 4, 1. If n = 5, the number of terms is 6 and the binomial coefficients are 1,5,10,10,5, 1.
In any term, the sum of the indices of a and b is equal to n.

15
 3a 
Illustration 2. Find the co-efficient of x24 in  x 2 +  .
 x 

Solution: The general term ((r + 1)th term) is

15 r
 2 3a  2 15–r 
3a 
 x + x  = Cr(x )
15
 x 
   
3r ar
= 15Cr x30–2r r = 15Cr 3rar x30–3r
x
If this term contains x24, then 30–3r = 24
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 3r = 6  r = 2.
Therefore, the co-efficient of x 24 = 15C2 9a2.

Illustration 3. If the binomial co-efficients of the (2r + 4) th term and the (r – 2)th term in the
expansion of (1 + x)18 are equal, find the value r.

Solution: The co-efficient of (2r + 4)th term in (1 + x)18 = 18C

2r+3 and the co-efficient of (r –
2)th term = 18Cr–3, so that 18C2r+3 = 18Cr–3
 2r + 3 + r – 3 = 18  3r = 18  r = 6.

n
 1
Illustration 4. If the 4th term in the expansion of  px +  is independent of x, find the value of
 x
n. Also calculate p if the 4th term is 5/2.

n −3
 1
Solution: Here T4 = T3+1 = nC3(px)3   = nC3 p3 x6–n.
x
T4 is independent of x  6 – n = 0 or n = 6.
Now (given), T4 = 5/2  6C3 . p3 = 5/2
5 1 1 1
 p3 = . 6 = p= .
2 C3 8 2

Middle term
(i) When n is even
th
n 
Middle term of the expansion is the  + 1 term
2 
i.e. Cn/2 a b in the expansion of (a + b)n.
n n/2 n/2

(ii) When n is odd

th th
 n + 1 n + 3
Middle terms of the expansion are the   term and the   term.
 2   2 
n+1 n-1 n-1 n+1
n n
These are given by, C n−1 a 2 b2 and C n+1 a2 b 2 in the expansion of (a + b)n.
2 2

e.g. middle term in the expansion of (1 + x) 4 and (1 + x)5:

Expansion of (1 + x)4 have 5 terms, so that third term is the middle term, which is the
th
4 
 2 + 1 term.
 
th
 5 + 1
Expansion of (1 + x)5 have 6 terms, so that 3rd and 4th are the middle terms, which are the  
 2 
th
5 +3
and the   terms.
 2 
Note:
◼ The binomial co-efficients in the expansion of (a+x) n, equidistant from the beginning and
from the end are equal.

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◼ The binomial coefficient of the rth term from the end = binomial coefficient of the
(n – r +2)th term from the beginning.
◼ If there are two middle terms, then the binomial co-efficients of two middle terms will be
equal and those two co-efficients will be greatest.

Illustration 5. Find the middle term in the expression of ( 1 – 2x + x2)n.

Solution: We have (1 – 2x + x2)n = [ (1 – x)2]n = (1 – x)2n.
Here 2n is an even integer
 2n 
  + 1 th i.e. ( n+1) th term will be the middle term.
 2 
( 2n ) !
Now (n + 1)th term in (1 – x)2n = 2nCn (1)2n-n(–x)n = 2nCn(–x)n = ( − x )n .
n!n!

Greatest Binomial Coefficient

To determine the greatest coefficient in the binomial expansion, (1 + x) n, when n is a positive
integer.
T  C n −r +1 n +1
Coefficient of  r +1  = r = = −1
 r 
T C r −1 r r
Now the (r + 1)th binomial coefficient will be greater than the rth binomial coefficient when
Tr+1  Tr
n +1 n +1
 − 1  1 r. .....(1)
r 2
But r must be an integer, and therefore when n is even, the greatest binomial coefficient is given
n
by the greatest value of r, consistent with (1) i.e., r = and hence the greatest binomial coefficient
2
is nCn/2.
Similarly if n be odd, the greatest binomial coefficient is given when,
n −1 n+1
r= or and the coefficient itself will be n C n +1 or n C n −1 .
2 2
2 2

To determine the numerically greatest term in the expansion of (a + x) n,when n is a positive integer.
Consider
Tr +1 n
Cr an−r xr n
Cr x n −r +1 x n +1 x
= n n − r +1 r −1
= n
=  = −1 .
Tr Cr −1a x Cr −1 a r a r a
n + 1  x n+1 a n +1
Thus |Tr + 1| > |Tr| if  − 1  1. i.e,  1+   r . …. (2)
 r a r x a
1+
x
 n+1 
Note:  − 1 must be positive since n > r .
 r 
Thus Tr+1 will be the greatest term, if r has the greatest value as per relation (2).

Illustration 6. Find the greatest term in the expansion of (2 + 3x)9 if x = 3/2.

Tr +1  n − r + 1   3x   10-r   3x   3
Solution: Here =   2  =  r   2  ,  where x= 2 
Tr  r         
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 10-r   3  10-r 9 90-9r

=    (3 / 2) =  = .
 r  2 r 4 4r

Therefore Tr+1  Tr if, 90 - 9r  4r

90
 90  13r  r  and
13
r being an integer, r = 6.
313 . 7
Hence Tr+1 = T7 = T6+1 = 9C6 (2)3 (3x)6 = .
2
10
 3 
Illustration 7. Given that the 4th term in the expansion of  2 + x  has the maximum numerical
 8 
value, find the range of values of x for which this will be true.

10 10
 3   3 
Solution: Given 4th term in  2 + x  = 210  1 + x , is numerically greatest
 8   16 
T T
 4  1 and 5  1
T3 T4
10 10
C3 3 C4 3
 10
. x  1 and 10
. x 1
C2 16 C3 16
64
 |x|  2 and |x| 
21
 64   64 
 x  − , −2  2, .
 21   21 

Particular Cases
We have, (a + x)n = an + nC1 an–1 x+ nC2 an–2 x2 + …. + nCr an–r xr + ….+ xn. …. (1)
(i) Putting x = –x in (1) , we get
(a –x)n = an – nC1 an–1x+ nC2 an–2x2 – nC3 an–3 x3 + ….+ (–1)r nCr an–rxr+….+ (–1)n xn.
(ii) Putting a = 1 in (1), we get,
(1+ x)n = nC0 +nC1x + nC2x2 + …. + nCr xr +…..+ nCnxn. …. (A)
(iii) Putting a = 1, x = –x in (1), we get
(1– x)n = nC0 – nC1 x + nC2 x2 – nC3x3+….(–1)r nCr xr +….(–1)n nCnxn. ….(B)

Tips to Remember

Tr +1 n − r + 1 x
(a) =  for the binomial expansion of (a + x)n.
Tr r a

(b) (n + 1)C = nCr + nCr-1

r

(c) When n is even,

(x + a)n + (x – a)n = 2(xn + nC2 xn-2 a2 + nC4 xn-4 a4 + .... +nCn an).

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When n is odd,
(x + a)n + (x – a)n = 2(xn + nC2 xn-2 a2+ ... +nCn-1 x an-1).
When n is even,
(x + a)n – (x – a)n = 2(nC1 xn-1a + nC 3 xn-3a3 + ... +nCn-1 x an-1).
When n is odd
(x + a)n – (x – a)n = 2(nC1xn-1a + nC3xn-3 a3 + ... +nCn an).

Exercise 1.
10
 1
i) Find the 7th term of  3 x 2 −  .
 3
ii) Find the term independent of x in the expansion of
10
 x 3   2 3 
25

(a)  + 
2 
(b)  2 x − 
 3 2x   x3 
iii) Find the middle term in the expansion of
6 7
 2x 3   2 1
(a)  −  (b)  2 x − 
 3 2x   x
iv) Find the term which does not contain irrational expression in the expansion of
( )
24
5
3+72 .
v) Find the numerically greatest term in the expansion of (3 – 2x)9 when x = 1.

Properties of Binomial Coefficients

For the sake of convenience, the coefficients nCo, nC1,....nCr,…..nCn are usually denoted by Co, C1,...
Cr ... Cn respectively.
Put x = 1 in (A) and get, 2n = Co + C1 +......+Cn
or Co + C1+ ..... + Cn = 2n. ....(D)
Also putting x = – 1 in (A) we get,
0 = Co – C1 + C2 – C3 +.......
 C0 + C2 + C4 + ........= C1 + C3 + C5 +….
But from (D) Co + C2 + C4 + ....... + C1 + C3 + C5 +.......=2n.
Hence Co + C2 + C4 + ......... = C1 + C3 + C5 +.......= 2n-1.
For x = 2, (A) gives nC0 + 2 nC1 + 22 nC2 + … + 2n nCn = 3n .

Some Important results

Differentiating (1 + x)n = C0 + C1x + C2x2 + ......+ Cnxn , on both sides we have,
n(1 + x)n-1 = C1 + 2C2x + 3C3x2 +.......+ nCnxn-1. ....(E)
Put x = 1in (E) so that n2n-1 = C1 + 2C2 + 3C3 +.......+ nC n.
Put x = –1 in (E) so that 0 = C1 – 2C2 + …. + (–1)n-1 nCn.
Integrating (1 + x)n , we have,

(1 + x )n+1 C1x 2 C2 x 3 C xn+1

+ C = C0 x + + + ......... + n (where C is a constant).
n +1 2 3 n +1
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1
For x = 0, we get C = – .
(n + 1)
(1 + x)n+1 − 1 C x2 C x3 C xn+1
Therefore = C0 x + 1 + 2 + ....... + n . ....(F)
n +1 2 3 n +1
2n+1 − 1 C C
Put x = 1 in (F) and get, = C0 + 1 + ....... + n .
n +1 2 n +1
1 C C
Put x = -1 in (F) and get, = C0 − 1 + 2 − ...........
n +1 2 3
3n+1 − 1 22 23 2n+1
Put x =2 in (F) and get, = 2 C0 + C1 + C2 + ....... + Cn .
n +1 2 3 n +1
Problems related to Series of Binomial coefficients in which each term is a product of an
integer and a binomial coefficient, i.e. in the form k.nCr.

n
Illustration 8. If (1+x)n = C x
r =0
r
r
then prove that C1 + 2C2 + 3C3+ . . . + nCn= n2n-1.

Solution: Method (i): rth term of the given series

rth term of the given series, tr = r nCr
n n-1 n
 tr = r   Cr-1 = n  n-1Cr-1 ( because nCr = . n-1Cr-1 ).
r r
n n
Sum of the series = t
r =1
r = n r =1
n −1
Cr −1 .

Put x =1 in the expansion of (1 + x)n-1, so that

( n −1
C0 + n −1
C1 + .....+ n −1
)
Cn−1 = 2n-1
n n
 t
r =1
r =n  r =1
n −1
Cr −1 = n .2n−1 .

Method (ii) By calculus

We have (1+ x)n = C0 + C1x + C2 x2 + .... + Cnxn. ….(1)
Differentiating (1) w.r.t. x, we get
n(1 + x)n-1 = C1 +2C2x + 3C 3 x2 + .... + n Cnxn-1. ….(2)
Putting x = 1 in (2), we have
n 2n-1 = C1 + 2C2 + .... + nC n.

n
Illustration 9. If (1+ x)n = C x
r =0
r
r
then prove that C0 +2.C1 + 3.C2 + ….+(n+1)C n = 2n–1(n+2).

Solution: Method (i): rth term of the given series

rth term of the given series, tr = r. nCr–1 = [(r –1) +1] .nCr–1
n n-1
= (r – 1) nCr–1 + nCr–1 = n. n–1Cr–2 + nCr–1 (because nCr-1 = . Cr-2).
r −1
n +1 n +1 n +1
Sum of the series = t
r =1
r =n 
r =1
n −1
Cr −2 + 
r =1
n
Cr −1

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= n[n–1C0 + n–1C1 + ….+n–1C n–1 ] + [nC0 + nC1 + …+nCn]

= n.2n–1 + 2n = 2n–1(n + 2).
Method (ii) by Calculus.
We have (1 + x)n = C0 + C1x + C2x2 + ….+C nxn. ….(1)
Multiplying (1) with x, we get
x(1 + x)n = C0x + C1x2 + C2x3 + ….+Cnxn+1. ….(2)
Differentiating (2) w.r.t. x, we have
(1 + x)n + n(1+ x)n–1x = C0 + 2C1x + 3C2x2 + ….+ (n + 1)Cnxn ….(3)
Putting x = 1 in (3), we get
2n + n. 2n–1 = C0 + 2C1 + 3C 2 + ….+ (n + 1)Cn
 C0 + 2C1 + 3C2 + ….+ (n + 1)C n
= 2n–1(n + 2).
Problems related to series of Binomial Coefficient in which each term is Binomial coefficient
divided by an integer, i.e. in the form of nCr / k.

2 n +1 − 1
n


C1 Cn
Illustration 10. If ( 1+ x)n = Cr x r , show that C0 + + + = .
r =0
2 n +1 n +1

Solution: Method (i): rth term of the given series

n
Cr −1 n + 1 n 1 1
tr = = . Cr −1 . = .n+1 Cr .
r r n +1 n +1
n +1 n +1 n +1

 
Cr
Sum of the series = tr =
r =1 r =1
n +1

=
1
n +1
( n +1
C1 +n+1 C2 + +n+1 Cn+1 )
2n+1 − 1
=
1 n+1
n +1
(C0 +n+1 C1 +n+1 C2 + +n+1 Cn+1 − n+1 C0 =
n +1
. )
Method (ii): By calculus
(1+x)n = C0 + C1x + C2x2 + . . + Cnxn .... (1)
Integrating both sides of (1) w.r.t. x between the limits 0 to x, we get
x
 (1 + x )n+1   C x2 C x3 C xn+1 
x
  = C0 x + 1 + 2 + + n 
 n +1  n + 1 
 0 
2 3
 0
n +1
(1 + x ) 1 C x2 C x3 Cn xn+1
 − = C0 x + 1 + 2 + + . ....(2)
n +1 n +1 2 3 n +1
2n+1 − 1 C C Cn
Substituting x = 1 in (2), we get = C0 + 1 + 2 + + .
n +1 2 3 n +1

Problem related to series of binomial coefficients in which each term is a product of two
binomial coefficients.

(a) If sum of lower suffices of binomial expansion in each term is the same
i.e. nC0 nCn + nC1 nCn–1 + nC2 nCn–2 + ….+ nCn nC0
i.e. 0 + n = 1 + (n – 1) = 2 + (n – 2) =….= n + 0.
Then the series represents the coefficient of x n in the multiplication of the following two
series
(1 + x)n = C0 + C1x + C2x2 + . . + Cnxn
and (x + 1)n = C0xn + C1xn–1 + C2xn–2 + . . + Cn.
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( 2n ) !
Illustration 11. Prove that Co2 + C12 + …. + Cn2 = .
n!n!

Solution: Since
(1 + x)n = C0 + C1x + C2 x2 + …. + Cnxn, …. (1)
(x + 1)n = C0xn + C1xn-1 + C2xn-2 + …. + Cn, ….. (2)
(C0 + C1x + C2x2 + …. + Cnxn) (C0 xn + C1xn-1 +C2 xn-2 + ….+ Cn)= (1 + x)2n.
Equating coefficient of xn, we get
2n!
C02 + C12 + C22 + ..... + Cn2 =2n Cn = .
n! n!

(b) If one series has constant lower suffices and other has varying lower suffices
Illustration 12. Prove that nC0.2nCn –nC12n –2Cn + nC2 .2n –4Cn – ….= 2n.

Solution: nC .2nC
0 n–nC1.2n –2Cn + nC2.2n –4Cn –……..
= co-efficient of xn in [nC0(1 + x)2n – nC1(1 + x)2n –2 + nC2(1 + x)2n –4 –……]
= co-efficient of xn in
[nC0((1 + x)2)n – nC1((1 + x)2)n –1 + nC2((1 + x)2)n –2 –……]
= co-efficient of xn in [(1 + x)2 – 1]n
= co-efficient of xn in (2x + x2)n = co-efficient of xn in xn (2 + x)n = 2n.

Exercise 2.
If (1 + x)n = C0 + C1x + C2 x2 + …. + Cnxn , prove that
i) 3Co – 8C1 + 13C2 – 18C3 + ... upto (n + 1) terms = 0.
C C C Cn 3 n +1 − 1
ii) 2C0 + 2 2 1 + 2 3 2 + 2 4 3 + ..... + 2 n +1 = .
2 3 4 n +1 n +1
( 2n ) !
iii) C0Cr+C1Cr+1+C2Cr+2+….+Cn-rCn=C0Cr+C1Cr+1+C2Cr+2+….+CrC0 = .
(n − r ) ! (n + r ) !
C1 C2 n Cn 1
iv) C0 − + − ..... + ( −1 ) = .
2 3 n +1 n +1
v) 3. nC1 + 7. nC2 + 11. nC3 + + ( 4n − 1).n Cn = ( 2n − 1)2 n + 1 .

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 Pinnacle Study Package -68
BT-10

Miscellaneous Exercise:
i) In the binomial expansion (1 + x)43, the coefficients of (2r + 1)th and the (r + 2)th terms
are equal. Find r.
ii) If three consecutive coefficients in the expansion of (1 + x)n are in the ratio 6 : 33 :
110, find n.
iii) Find the ratio of the coefficient of x10 in (1 − x2)10 and the term independent of x
in (2 + 2/x)10.
iv) Find the greatest term in magnitude, in the expansion of (3 − x)7 when x = 1.
v) Find the coefficient of x10 in he expansion of (x2 − 2)11.
vi) Show that the coefficient of the middle term in the expansion of (1 + x) 2n is equal to
the sum of the coefficients of the two middle terms in the expansion of (1 + x)2n−1.
vii) If C1, C2, C3, C4, …. are the coefficient of the 2nd, 3rd, 4th and 5th terms in the expansion
C1 C3 2C2
of (1 + x)n, prove that + = .
C1 + C2 C3 + C4 C2 + C3
viii) Find the values of m such that the coefficient of x2 in the expansion of (1 − x)m is 3.
ix) Find the value of (0.99)5 correct to three decimal places.
x) Prove that a C0 + (a + d)C1 + (a + 2d)C2 + … (a + nd)Cn = (2a + nd)2n−1, where C0, C 1,
C2,… Cn are the binomial coefficient in the expansion of (1 + x)n.

ANSWERS TO EXERCISES

Exercise 1.
70 8
i) x
3

5 25! 15 10
ii) (a). (b). 2 3
12 10! 5!

iii) (a). –20 (b). −560.x 5 , 280x 2

24! 2 2
iv) 3 2 v) 4th and 5th i.e. 489888
10!14!

Miscellaneous Exercise:

i) 1, 14 ii) 12
iv) 7(36) v) 29568
viii) 3, − 2 x) 16
 Pinnacle Study Package -68
-BT-11

ASSIGNMENT PROBLEMS

Level − 0

1 1 1 1
1. Prove that + + + ... = 2n−1 .
n! 2! ( n − 2 )! 4! (n − 4 )! n!

2. Find the 10th term in the expansion of

12
 1
(i)  2x 2 +  (ii) (x – y2)15.
 x
10
 1 
3. Write the middle term in the expansion of  x −  .
 2y 
4. If the middle term in the expansion of (1 + x)2n is the greatest, prove that
n n +1
x .
n +1 n
9
 x3 2 
5. Find the fourth term from the end in the expansion of  −  .
 2 x2
 

6. Find the term independent of x in the expansion of

10
 1
9

15
3   x 3 
(i)  x 2 +  , (ii)  x 3 − 3  , (iii)  +  .
 x  x   3 2x 2
 
7. In the binomial expansion of (1 + y) n, where n is a natural number, the coefficients of the
5th, 6th and 7th terms are in A.P. Find n.
8. How many terms are there in the binomial expansions

( )
10 5 6
(ii) ( 2x + 3y )  (iii) ( x − y ) 
3 1/ 3
(i) 2x + 5y
   

( ) ( )
6 6
9. Expand (a + b)6 − (a − b)6. Hence find the value of 2 +1 − 2 −1 .

9
 1 
10. Find the coefficient of x9 and the term independent of x in the expansion of  x 2 −  .
 3x 
n
 1
11. Find the term independent of x in (1 + x)m  1 +  .
 x

12. Prove that 62n − 35n − 1 is divisible by 1225.

13. Prove that (106)85 − (85)106 is divisible by 7.

14. If the coefficients of rth, (r + 1)th and (r + 2)th terms in the expansion of (1 + a)n are in A.P.,
prove that n2 − 21n + 98 = 0. Find n.

15. If the three successive coefficients in the expansion of (1 + x)n are 36, 84 and 126, find n.

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 Pinnacle Study Package -68
BT-12

ANSWERS TO ASSIGNMENT PROBLEMS

Level − 0

1. n = 4, a = 2

1760
2. (i) 3
(ii) − 5005x6y18
x

−63x 5 672
3. 5.
8y 5
x3

6. (i) 84 (ii) − 15 C6  39 (iii) 5/12

7. n=7

8. (i) 6, (ii) 16, (iii) 3

9. 140 2

−28 29 m+nC
10. , 11. n
9 243

14. n = 7, 14 15. n=9

 Pinnacle Study Package -68
-BT-13

SOLVED PROBLEMS

Subjective:

9
3 1 
Problem 1. Find the term independent of x in the expansion of (1+x+2x 3 )  x 2 −  .
2 3x 

9
3 1 
Solution: (1 + x + 2x3)  x 2 − 
 2 3x 

 3 
9
3 
8 7 2
+ 9C2  x 2    
1 3 1
= (1 + x + 2x 3 )  x 2  −9 C1  x 2  
 2   2  3x  2   3x 
6 3 5 4 4 5
− 9C3  x 2     +9 C4  x 2    −9C5  x 2   
3 1 3 1 3 1
 2   3x   2   3x   2   3x 
3
3   1 
6 2
3   1 
7 
+9 C6  x 2     − 9C7  x 2    + ..... .
 2   3x   2   3x  
3 2 7
Hence term independent of x = 9C6    3  1
3 1
− 2.9 C7     
2 3 6
2 3
1 1 9! 1 9! 1
= 9 C6  − 2.9 C7 =  − 2 
8  27 4.3 5 6! 3! 8  27 7! 2! 4.243
789 1 89 1 17
=  − 2.  = .
6 8.27 2 4.243 54

Problem 2. If a1, a2, a3, a4 are the coefficients of any four consecutive terms in the expansion
a1 a3 2a2
of (1+x)n , prove that + = .
a1 + a2 a3 + a4 a2 + a3

Solution: Let a1,a2, a3,a4 be the coefficients of the consecutive terms tr+1, tr+2, tr+3 and tr+4 in
the expansion of (1+x)n so that, a1= nCr , a2 = nCr+1, a3 = nCr+2 , a4 = nCr+3
 a1 + a2 = nCr + nCr+1 = n+1Cr+1,
a2 + a3 = nCr+1 + nCr+2 = n+1Cr+2
and a3 + a4 = nCr+2 + nCr+3 = n+1Cr+3.
n
a1 C n! (r + 1)!(n − r)! r + 1
Now, = n+1 r = = .
a1 + a2 Cr +1 r !(n − r)! (n + 1)! n +1
a2 r+2 a3 r+3
Similarly, = , = .
a 2 + a3 n + 1 a3 + a 4 n +1
a1 a3 r +1 r+3 2 (r + 2) 2a2
Hence + = + = = .
a1 + a2 a3 + a4 n +1 n +1 n +1 a 2 + a3

Problem 3. Let n be a positive integer. If the co-efficients of 2nd, 3rd and 4th terms in the
expansion of (1+x)n are in A.P., then find the value of n.

Solution: Here t2 = nC1 x, t3 = nC2 x2, t4 = nC3 x3.

Since the coefficients are in A.P,
2 nC2 = nC1 + nC3

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 Pinnacle Study Package -68
BT-14

n ( n − 1) n (n − 1)(n − 2) n(6 + n2 − 3n + 2)
2 = n+  n(n–1) =
2 ! 3 ! 6
 6(n – 1) = n – 3n + 8  n – 9n + 14 = 0  n = 7 as n can not be 2.
2 2

5 5
 1   1 
Problem 4. Find the degree of the expression  x + ( x 3 − 1) 2  +  x − ( x 3 − 1) 2  .
   

5 5
 1  1
Solution: Since  x + (x 3 − 1) 2  +  x − (x 3 − 1) 2 
   
  1 2  1 4 
3  3  
= 2 x + 10x (x − 1)  + 5x (x − 1)  
 5 2 3 2
 
     

= 2  x 5 + 10x 3 (x 3 − 1) + 5x(x 3 − 1)2  , degree is 7.
 

Problem 5. Find the larger of 9950 + 100 50 and 10150 .

Solution: We have (101)50 = (100 + 1)50

50.49
= 10050 + 50.10049 + 10048 + ….
1.2
and 9950 = (100–1)50
50.49
= 10050 – 50. 10049 + . 10048 – …..
1.2
Subtracting, we get
 50.49.48 
(101)50 – 9950 = 2 50.100 49 + 100 47 + ....
 1.2.3 
 50.49.48 
= 10050 + 2 100 47 + ..... > 10050.
 1.2.3 
Hence, (101)50 > 9950 + 10050.

Problem 6. Using Binomial Theorem show that 32n+2– 8n– 9 is divisible by 64,  nN.

Solution: The given expression 32n+2 – 8n – 9 = 32(32)n – 8n – 9

= 9(1 + 8)n–8n – 9
= 9 1 +n C1 8 +n C2 .82 + ..... + 8n  − 8n − 9
= 64 n + 9(nC282 + nC3.83 + …. +8n)
= 64 [n+9(nC2+nC3.8 + …. + 8n-2)]
= 64K, where K is an integer.
Hence 32n+2 – 8n – 9 is divisible by 64.

Problem 7. Find the remainder when 599 is divided by 13.

Solution: We can write 599 = 5.2549 = 5(26 − 1)49

= 5(C02649 − C12648 + C22647 − C32646 + + C48 26 − C49)
= [5 (C02649 − C12648 + + C48 26 ) − 13] + 8.
Hence on dividing by 13, remainder is 8.
 Pinnacle Study Package -68
-BT-15

Problem 8. If n be any positive integer, show that the integral part of (7 + 4 3 )n is an odd
number. Also if (7 + 4 3 )n = I + f where I is a positive integer and f is a proper
fraction, show that (1 – f) (I + f) = 1.

( )
n
Solution: Let 7 + 4 3 = I + f where I is a +ve integer and f is a proper fraction.
Clearly 0 < 7 − 4 3 < 1  0 < (7 − 4 3 )n < 1.
Let (7 − 4 3 )n = f1 where 0 < f1 < 1.
Now I + f = 7n + nC1 7n−1(4 3 ) + nC2 7n−2 (4 3 )2 + + (4 3 )n, …. (1)
and f1 = 7n − nC1 7n−1(4 3 ) + nC2 7n−2 (4 3 )2 +(−4 3 )n. …. (2)
Adding (1) and (2), we get
 
( )
2
I + f + f1 = 2 7n +n C2 7n−2 4 3 +  = an even integer
 
 f + f1 = even integer − integer = an integer.
Since 0 < f < 1, 0 < f1 < 1,
0 < f + f1 < 2  f1 + f = 1
 I = even integer – 1 = Odd integer.
Now, I + f = (7 + 4 3 )n and f1 = (7 − 4 3 )n
 (I + f) (1 − f) = (49 − 48)n = 1.
n
Problem 9. If p + q = 1, then show that r
r =0
2
Cr p r q n −r = npq + n 2 p2 .

Solution: Here tr = r2 nCr prqn–r = r.n. n–1Cr–1 pr qn–r

= n[ (r –1) + 1] n–1Cr–1 pr qn–r
= n[ (r –1) n–1Cr–1 + n–1Cr–1] pr qn–r
= n [ (n –1) n–2Cr–2 + n–1Cr–1] pr qn–r
= n (n –1) n–2Cr–2 pr qn–r + n .n–1Cr–1 pr qn–r
= n (n –1) p2 n–2Cr–2 pr–2 qn–r + n p n–1Cr–1 pr–1 qn–r.
n
Now sum S = t
r =0
r = n(n –1)p2 [n–2C0 qn–2 + n–2C1 pqn–3 + ….+ n–2Cn–2pn–2]

+ np[n–1C0 qn–1 + n–1C1 pqn–2 + ….+ n–1Cn–1 pn–1]

= n(n –1)p2
(p+ q)n–2
+ np( p + q)n–1
= n(n –1)p + np( p +q =1)
2

= n2 p2 – np2 + np = n2p2 + np(1 – p) = n2p2 + npq.

Alternative Solution:
n
We know that (1 + x)n = r =0
n
Cr xr .

Differentiate both sides w.r.t x and get

n n
n(1 + x)n – 1 = r
r =0
n
Cr xr −1  nx(1 + x)n – 1 = r
r =0
n
Cr xr .

Differentiate both sides again, w.r.t x. We get

n
n(1 + x)n – 1 + n(n – 1)x(1 + x)n – 2 = r
r =0
2 n
Cr xr −1

n
 nx(1 + x)n – 1 + n(n – 1)x2(1 + x)n – 2 = r
r =0
2 n
Cr xr .

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 Pinnacle Study Package -68
BT-16

p
Write x = in this equation, so that
q
np(p + q)n−1 n(n − 1)p2 (p + q)n−2
n
pr
qn
+
qn
=  r =0
r2 n
Cr
qr
n
 np + n(n – 1)p2 = r
r =0
2 n
Cr pr qn−r (as p + q = 1)

n
 r
r =0
2 n
Cr pr qn−r = n(1 – p)p + n2p2 = npq + n2p2.

2n 2n
Problem 10. If 
r =0
ar ( x − 2 ) =
r
b
r =0
r ( x − 3 )r and ak = 1 for all k  n,

then show that bn = 2n+1Cn +1.

2n 2n
Solution: Writing x - 3 = y, we get a
r =0
r (1 + y )r = b y
r =0
r
r
. We have to find the coefficient of

yn in the left hand side as the coefficient of yn in the right hand side is bn. And we
will start getting coefficient of y n only when r  n in the left hand side.
Coefficient of yn in (1 + y)n + (1 + y)n + 1 + ... + (1 + y)2n = bn.
 (1 + y )n+1 − 1
 
Coefficient of y in (1 + y) 
n n
 = bn
 y 

 Coefficient of yn + 1 in (1 + y ) 2n +1
− (1 + y )
n
 =b n  2n+1Cn+1 = bn.
 Pinnacle Study Package -68
-BT-17

Objective:

Problem 1. The total number of terms in the expansion of (x + a)100 + (x – a)100 after
simplification is
(A) 50 (B) 202
(C) 51 (D) none of these

Solution: Total number of terms in the expansion of (x + a) 100 is 101, of which 50 terms get
cancelled.
Hence (C) is the correct answer.

Problem 2. The co-efficient of y in the expansion of (y2 +c/y)5 is

(A) 10 c3 (B) 20 c2
(C) 10 c (D) 20 c

5 −r
c
( )
r
Solution: General term in the expansion = Cr y 5 2
  = 5 Cr y3r −5 c 5 −r
y
For coefficient of y, put r = 2.
Hence (A) is the correct answer.

9
 1
Problem 3. The term independent of x in  x 2 −  is
 x
(A) 1 (B) –1
(C) 48 (D) 84

r 9 −r
 −1 
Solution: General term in the expansion is 9
( )
Cr x 2  x 
 
9 −r
= 9 Cr x 3r −9 ( −1)
For constant term, put r= 3.
Hence (D) is the correct answer.

n n
Problem 4. If (1 + x)n = 
r =0
n
Cr x r , then 
r =m
r
Cm is equal to

(A) n + 1Cm (B) n + 1Cm + 1

(C) n + 2Cm + 1 (D) none of these

Solution: rC is coefficient of xm in (1 + x)r

m
n
 
r =m
r
Cm is coefficient of xm in (1 + x)m + (1 + x)m + 1 + …… + (1 + x)n

i.e. coefficient of xm in

(1 + x)m (1 + x)n+1−m − 1 
x
or coefficient of xm + 1 in (1 + x)n + 1 = n + 1Cm + 1.
Hence (B) is the correct answer.
n
 x
Problem 5. If the co-efficients of x7 and x8 in  2 +  are equal, then n is
 3
(A) 56 (B) 55
(C) 45 (D) 15

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 Pinnacle Study Package -68
BT-18

2n−7 2n−8
Solution: Here n C7 = n
 n = 55.
C8
37 38
Hence (B) is the correct answer.

2n
 1 
Problem 6. If xm occurs in the expansion of x + 2  , then the co-efficient of xm is
 x 
( 2n ) ! ( 2n ) !
(A) (B)
 2n − m   4n + m  n ! ( 2n − m ) !
  !   !
 3   3 
( 2n ) ! 3 ! 3 !
(C) (D) none of these
( 2n − m ) !

1
Solution: General term in the expansion is 2n
Cr x 2n−r 2r
= 2n
Cr x 2n−3r
x
2n − m
For xm, 2n − 3r = m  r =
3
so that the coefficient of xm is 2n C 2n−m .
3
Hence (A) is the correct answer.

Problem 7. The positive integer which is just greater than (1+0.0001) 1000 is
(A) 3 (B) 4
(C) 5 (D) 2

Solution: The given expression on expansion gives

1000  999 −8
1 + 1000  10−4 + 10 + 1000 C3 10 −12 + …….
2
1 1 1 1 10
< 1+ + + + = =
10 100 1000 1 9
1−
10
so that the integer just greater than the given expression must be 2.
Hence (D) is the correct answer.

Problem 8. Given the integers r>1, n> 2, and co-efficients of (3r) th and (r+2)th term in the
binomial expansion of (1+x)2n are equal, then
(A) n = 2r (B) n =3r
(C) n = 2r+1 (D) none of these

Solution: Coefficients of (3r)th and (r + 2)th terms will be 2nC3r−1 and 2nCr+1.
These are equal
 (3r − 1) + (r + 1) = 2n n = 2r.
Hence (D) is the correct answer.

Problem 9. The co-efficient of the middle term in the expansion of (1+x)2n is

1.3.5. ....( 2n − 1) n
(A) 2 nCn (B) 2
n !
(C) 2. 6. …. (4n – 2) (D) none of these
 Pinnacle Study Package -68
-BT-19

2n!
Solution: Coefficient will be 2n
Cn =
(n!)2
1.3.5 (2n − 1). 2.4.6.8 2n 1.3.5 (2N − 1) n
= 2
= 2 .
(n!) N!
Hence (B) is the correct answer.

Problem 10. If (1–x+x2)n=a0+a1x+a2x2 + …. + a2nx2n then a0 + a2 + a4 + …. +a2n equals

3n + 1 3n − 1
(A) (B)
2 2
1−3 n
1
(C) (D) 3 n +
2 2

Solution: Put x = 1  1 = a0 + a1 + a2 + + a2n.

Put x = −1  3n = a0 − a1 + a2 − a3 + + a2n.
Adding, 3n + 1 = 2 (a0 + a2 + a4 + + a2n).
Hence (A) is the correct answer.

Problem 11. The co-efficients of xp and xq (p and q are positive integers) in the expansion of (1
+ x)p+q are
(A) equal (B) equal with opposite signs
(C) reciprocals to each other (D) none of these

p+q p+q
Solution: The coefficients of xp and xq are Cp and Cq both of which will be equal.
Hence (A) is the correct answer.

( )
10
Problem 12. The sum of the rational terms in the expansion of 2 + 31 / 5 is
(A) 46 (B) 42
(C) 41 (D) none of these
10 10 −r

( ) 
10
Solution: The given expansion 2 + 31/ 5 = 10
C2 2 2 3r / 5 .
r =0
The rational terms correspond to r = 0, r = 10. Hence the sum of rational terms =
10C 25 + 10C 2
0 10 3 = 32 + 9 = 41.
Hence (C) is the correct answer.

Problem 13. The value of (n+2)C02n+1 – (n+1)C12n+ nC 22n-1+ . . . is equal to

(A) 4n (B) 4
(C) 2n+4 (D) 4(1+n)
Solution: Since (x-1)n = C0xn – C1xn-1+ . . .
x2(x-1)n = C0xn+2 – C1xn+1+ . . .
Differentiating w.r.t. x we get
2x(x-1)n - x2 n(x-1)n -1= (n+2)C0xn+1 – (n+1) C1xn+ . . .
Put x=2 and the result is 4(n+1).
Hence (D) is the correct answer.

Problem 14. The number of terms in the expansion of (a + b + c)n, where nN, is
( n + 1)( n + 2 )
(A) (B) n+1
2
(C) n+2 (D) (n+1)n

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 Pinnacle Study Package -68
BT-20

Solution: (a + (b+c))n = an +nC1 an−1(b + c)1+nC2an−2 (b + c)2 + + nCn (b + c)n.

Further expanding each term of R.H.S.,
first term on expansion gives one term,
second term on expansion gives two terms,
third term on expansion gives three terms and so on
(n + 1) (n + 2)
 Total no. of terms = 1 + 2 + 3 + + (n + 1) = .
2
Hence (A) is the correct answer.

Problem 15. The value of the expression

n (1 + x ) n ( n − 1 ) (1 + 2x ) n( n − 1)( n − 2 ) (1 + 3x )
1− . + − + ........ is
1 1 + nx 1.2 (1 + nx )2 1.2.3 (1 + nx )3
(A) 2 (B) 1
(C) 3 (D) 0

Solution: The expression can be divided into two parts as

 n 1 n(n − 1) 1 n(n − 1)(n − 2) 
 1 − + − + 
 1 (1 + nx) 2 (1 + nx)2
1  2  3 (1 + nx)3 
 −nx n(n − 1)x n(n − 1)(n − 2) x 
+  + − + 
 1 + nx (1 + nx)2 1.2 (1 + nx)3
 
nx  
n
 1  (n − 1) (n − 1)(n − 2)
= 1 −  − 1 − + + 
 1 + nx  1 + nx  1(1 + nx) 1.2(1 + nx)2 
n n −1 n n −1
 nx  nx  1   nx   nx   nx 
=   − 1− =  − = 0.
 1 + nx 

1 + nx  1 + nx  
 1 + nx 
  
 1 + nx   1 + nx 
Hence (D) is the correct answer.
 Pinnacle Study Package -68
-BT-21

ASSIGNMENT PROBLEMS

Subjective:

Level – I

8
1 
1(i). Find the term independent of x in the expansion of  x1/ 3 + x −1/ 5  .
 2 
11 11
 1   1 
(ii). Find the coefficient of x 7 in  ax 2 + -7
 and of x in  ax- 2  and find the relation
 bx   bx 
between a and b so that their coefficients are equal.

(iii). Find the middle term in the expansion of

9
a 
12
 x3 
(a)  x + bx  , (b)  3x −  .
   6 

2. Determine the value of x in the expression (x + xlog10 x )5 if the third term in the expansion is
10,00,000.
n
 1 
Find n in the binomial  3 2 +
3 
3. , if the ratio of 7th term from the beginning to 7th term
 3 
from the end is 1/6.
4. If x4r occurs in the expansion of (x + 1/x 2)4n, prove that its coefficient is
( 4n ) !
.
4  4 
 3 (n − r)  !  3 (2n + r)  !
   

5. Prove that the coefficient of (r + 1)th term in the expansion of (1 + x)n+1 is equal to the sum
of the coefficient of rth and (r + 1)th terms in the expansion of (1 + x)n.
6. In the expansion of (x+a)n, if the sum of odd terms be P, and sum of even terms be Q,
prove that
(a) P2 − Q2 = (x 2 − a2 )n , (b) 4PQ = (x + a)2n − (x − a)2n .
21
  a  b 
Which term in the expansion of the binomial  3  +  contains same power on
  b 
7.

3
a 
‘a’ and ‘b’.

8. Prove that 10 ( 10 + 1)
100
− ( )
10 − 1
100
 is a whole number.
20
 1 
9 (i). Find the value of greatest term in the expansion of 3 1 +  .
 3

(ii). If x = 1/3, find the greatest term in the expansion of (1 + 4x)8.

C1 C C C n(n + 1)
10. Prove that + 2 2 + 3 3 + …. + n n = .
C0 C1 C2 Cn-1 2

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 Pinnacle Study Package -68
BT-22

Level – II

1. If n is an in integer greater than 1, prove that

a – C1(a – 1) + C2(a – 2) + ….. + (–1)n Cn (a – n) = 0.

2. If (1 + x)15 = C0 +C1x +C2x2 +......+C 15x15, find the value of

C2 + 2C3 + 3C4 +......+14C15.

C0 C2 C 4 2n
3. Prove that + + + ...... = .
1 3 5 n +1

C1 C3 C5 2n − 1
4. Prove that + + + ...... = .
2 4 6 n +1

5 (i). Show that (1 + x)n – nx – 1 is divisible by x 2 ,  n  N.

(ii). Show that 23n – 7n – 1 is divisible by 49. Hence show that 23n+3 – 7n – 8 is divisible by 49.

6. If a, b, c are three consecutive coefficients in the expansion of (1 + x) n then prove that

2ac + b(a + c)
n= .
b2 − ac

7. If (1+ x –2x2)6 = 1+ a1x+ a2x2 +….+a12x12, then prove that a2 + a4+ …+ a12 = 31.

n+ 4
8. Let (1 + x2)2 (1 + x)n = a x
k =0
k
k
. If a1, a2 and a3 are in A.P., find n.

2
n
 C 
9. If n is a positive integer and Ck = nC
k then find the value ofk =1
3
k  k  .
 Ck −1 

( ) ( ) ( )
6 6 6
10. Show that 2 +1 + 2 −1 = 198. Hence show that the integral part of 2 +1
is 197.
 Pinnacle Study Package -68
-BT-23

Objective:

Level – I

1. If the second, third and fourth terms in the expansion of (a+b) n are 135, 30 and 10/3
respectively, then the value of a is
(A) 3 (B) 4
(C) 5 (D) 7

2. If the binomial co-efficient of rth, (r+1)th and (r+2)th terms in the expansion of (1+x) 14 are
in A.P., then the value of r is
(A) 10 (B) 6
(C) 7 (D) 9

3. If (1 + ax)n = 1 + 8x + 24x2 + ….. then

(A) a= 3 (B) a = 4
(C) a= 2 (D) a = 5

4. If in the expansion of (1 + x)m(1 – x)n, the coefficients of x and x2 are 3 and –6 respectively,
then m is
(A) 6 (B) 9
(C) 12 (D) 24

5. The two successive terms in the expansion of (1+x)24 whose coefficients are in the ratio 1
: 4 are
(A) 3rd and 4th (B) 4th and 5th
(C) 5th and 6th (D) 6th and 7th
6. The sum nC0 +nC1 +nC2+…..+nCn is equal to
2.4.6......2n
(A) (B) nn
n !
(C) n! (D) 3n

7. The co-efficient of xk (0kn) in the expansion of E = 1+(1+x) +(1+x) 2+…..+(1+x)n is

(A)n+1Ck+1 (B) nCk
(C) n+1Cn–k–1 (D) n+1Cn-k-1
8. The ratio of the co-efficient of x15 to the term independent of x in [x2+2/x]15 is
(A) 12: 32 (B) 1:32
(C) 32 :12 (D) 32:1
n log3 8
 1   1 
9. The last term in the binomial expansion of  3 2 −  is  3  . Then the 5th term
 2  3. 9 
from the beginning is
(A) 4.10C6 (B) 2.10C4
1
(C) .10 C4 (D) none of these
2

10. The remainder when 22003 is divided by 17 is

(A) 1 (B) 2
(C) 8 (D) none of these

11. The co-efficient of x in the expansion of (1-2x3+3x5)[1+(1/x)]8 is

(A) 56 (B) 65
(C) 154 (D) 62

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 Pinnacle Study Package -68
BT-24

12. The greatest value of the term independent of x in the expansion of (xsin p +x –1cosp)10,
p R, is
10!
(A) 25 (B)
(5!)2
1 10!
(C) 5
. (D) none of these
2 (5!)2

13. The co-efficient of x5 in the expansion of (1+x 2)5 (1+x)4 is

(A) 40 (B) 50
(C) 30 (D) 60

100
14. The co-efficient of x53 in the expansion 
m =0
100
Cm (x − 3)100 −m 2m is

(A) 100C53 (B) - 100C53

(C) 65C53 (D) 100C65

15. The term independent of x in the expansion of (x+1/x) 2n is

1.3.5. .....(2n − 1).2n 1.3.5.....(2n − 1).2n
(A) (B)
n ! n ! n !
1.3.5. .... (2n − 1) 1.3.5. ....(2n − 1)
(C) (D)
n ! n ! n !
 Pinnacle Study Package -68
-BT-25

Level – II

1. In the binomial expansion of (a – b)n, n  5, the sum of the 5th and 6th terms is zero. Then
a
equals
b
n−4 n−5
(A) (B)
5 6
5 6
(C) (D)
n−4 n−5

8
 1 
2. If 6th term in the expansion of  8 / 3 + x 2 log10 x  is 5600, then x is equal to
x 
(A) 5 (B) 4
(C) 8 (D) 10

n
 1 
3. If in the expansion of  2x + x  , T3/T2 = 7 and the sum of the co-efficients of 2nd and 3rd
 4 
terms is 36, then the value of x is
(A) –1/3 (B) –1/2
(C) 1/3 (D) 1/2

4. The two consecutive terms in the expansion of (3+2x) 74 whose coefficients are equal is
(A) 30th and 31st term terms (B) 29th and 30th terms
st nd
(C) 31 and 32 terms (D) 28th and 29th terms
n n
5. If 
r =m
r
Cm = n+1Cm +1 , then  (n − r + 1) C
r =m
r
m is equal to

(A) n + 3Cm + 2 (B) n + 2Cm + 2

(C) n + 2Cm + 1 (D) none of these

6. In the expansion of [71/3+111/9] 6561 , the number of terms free from radicals is
(A) 730 (B) 715
(C) 725 (D) 750

( )
2m
7. The integer next above 3 +1 contains
(A) 2m+1 as a factor (B) 2m+2 as a factor
(C) 2m+3 as a factor (D) 2m as a factor

8. If 13Cris denoted by Cr, then the value of C1+C5+C7+C9+C11 is equal to

(A) 212-287 (B) 212-165
(C) 212-C3 (D) 212-C2-C13
9. Let n be an odd natural number greater than 1. Then the number of zeros at the end of
the sum 99n +1 is
(A) 3 (B) 4
(C) 2 (D) none of these
n
10. If x +y =1, then r
r =0
n
Cr xr yn−r equals

(A) 1 (B) n
(C) nx (D) ny

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 Pinnacle Study Package -68
BT-26

11. The co-efficient of x4 in the expansion of (1+x+x2+x3)11 is

(A) 330 (B) 990
(C) 900 (D) 895

12. The co-efficient of xn in (1+x+2x2+3x3+ …. + nxn)2 is

(A)
(
n n2 + 11 ) (B)
(
n n2 + 10 )
6 6

(C)
(
n n2 + 11 ) (D)
(
n n2 + 10 )
4 4

10
 k 
13. If the term independent of x in  x + 2  is 405, then value of k is
 x 
(A) 2 (B) –2
(C) 3 (D) none of these

14. The coefficient of x11 in the expansion of (2x 2 + x – 3)6 is

(A) 384 (B) 192
(C) 572 (D) 64

Pn+1
15. If Pn denotes the product of all the co-efficients in the expansion of (1+x) n, then is
Pn
equal to
(n + 2 )n (n + 1)n+1
(A) (B)
n ! (n + 1) !
(n + 1)n+1 (n + 1)n
(C) (D)
n ! (n + 1) !
 Pinnacle Study Package -68
-BT-27

ANSWERS TO ASSIGNMENT PROBLEMS

Subjective:

Level - I

11 a6 11 a5
1 (i). j r = 5, t6 = 7 (ii) C5 , C6
b5 b6

189 17 −21 19
(iii) (a) 7th term, T7 = 924 a6b6 (b) T5 = x , T6 = x
8 16

2. x = 10 or 10–5/2 3. n=9

25840
7. 10th term. 9 (i) t8 = (ii) r = 5
9

Level – II

2. 214.13 + 1 8. n = 2, 3, 4

n ( n + 1) (n + 2 )
2
9.
12

Objective:

Level – I

1. A 2. D
3. C 4. C
5. C 6. A
7. A 8. B
9. A 10. C
11. C 12. C
13. D 14. B
15. A

Level – II

1. B 2. D
3. A 4. A
5. B 6. A
7. A 8. A
9. C 10. C
11. B 12. A
13. C 14. B
15. B

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