JR-IPE-MATHS-IA

MATHEMATICAL INDUCTION
LONG ANSWERQUESTIONS:
SOLUTION
n+D'(n+2)
nterms -,VneN
Fooos upto 12
+2' +3)+.
***1. Show that l' +(1+2')+(1'
(Mar-09, 12, May-09,Mar-16)

+--+rt)=Mn+I)(n+2)
S(n)=I+{l+2')+(r+2+3}4---+(l+2+3 12 tneN
SolLet n(n+1)(2n+1)
12

(IM)
P+(f+2)+{ +2+3)+---+ 6
12

for n=1.
Step-I: To prove that S(n) is true
LHS = 6
6

RHS
12
(IM)
. LHS =RHS, S(1) is true.
Step-II: Assumethat S(n) is true for n=k.
k(k+1)(2k+1)_ k(k+1) (k+2)-....).(IM)
ie; +(e +2')+(> +2' +3)+---+ 6 12

Step-II: We have to prove that S(n) is true for n =k+1

Adding on both sides

(k+1)(k +2)%2k+3) in eq.(1),we get
6

P+(+2)+(P+2+$)+-- k(k+1)(2k
6
+1),(k+1)(k+2)(2k+3)
6

k(k+) (k+2),(k+1)(& +2)(2k +3)_(k+1)(k+2)| k(k2+1) +2k+3

12 6

(k+1)(k+2)| k+k+4k +6 (k+1)(k +2)) +Sk +6|

6 2 6

(k+)(k +2)(k+2)(k+3)
12
(k+1)(k+2j(k +3)_(k+1)(k +1+1)}(k+1+2)
12 12
S(k+1) is true (3M)

Bythe principle of finite mathematical induction S(n) is true. tne N (IM)

P P+2. P+2+3
at*,Show that t+
1 1+3 1+3+5
-upto nterms =
2r +9n+13 (March-05,07,2014)
24

NARAYANA Page.no.18
JR-IPE-MATHS-IA SOLUTIONS

= +
'+2,+2' +3' I+2' +3' +....tn'
I+345+.... n
(2n-1)2n +9n+
Sol: LetS(n) 1
t......+
1+3 1+3+5 24

L,l+2, I'+2'+3'
1+3 1+3+5
al2n'+9n +13|
1
[2+(o-192]
=,P+2,'+2+3' n'(n+1)14
1 I+3 1+3+5 n'
2r'+9n +13 4.

i'+2' +24
+3' t + 1 ) ? (IM)
I+3 I+3+5 4
-2r'
24
+9n +13]
1
Step-I: To prove that S(n) is true for n=1.
LHS=7+)(l+1)' 4
4 4 4
1

RHS-+9n+13]-2*9+ 13]=1
(IM)
: L.H.S =R.H.S, S(1) is true.

Step-I: Assumethat S(n) is true forn =k.

P+2° P+2 +3 +......+
(k +1y? k
2*' +9k +13 . . ) . (IM)
-+ +
1 1+3 1+3+5 4 24L
n=k+1
Step-III: We have toprove that S(n) is true for
(k+2)²
(k+1)® term = 4

Adding (k +1)" term on both sides of (1)

P P+2 1+2 +3? +.....+ (k+1, (k+2)
4 24
2k+9k +13|+ (k+2)?
4
1 1+3 1+3+5 4
+37k +24
6(k +4k +4) 2k +9k2 +13k +6k +24k+ 24 2k³+15k'
2 +9k +13k 24
24
24 24

+13|
+6k +6k +2+9k +18k +9+13k

-2(4' +34 +3k +)+94 +2k +)+13k +)]

24
+)+13] (3M)
-2k+1y' +9|k+1Ý +13k+1)]-2+9
24
+%k
S(k+1) is true. (IM)
offinite mathematical induction S(n) is true,Vn eN
.:.Bytheprinciple

Page.no.19
NARAYANA
 JR-IPE-ATHS-IA +---upto ntems = sOLWT
***3.S that Yne N4
Sol:
4.7 7.10
I.4,7, . . .arein A.P, a =l,d=3
,=0+(n-)d=1+(n-1)3 =1+ 3n-3 =31-2
3n +1

: h AP.
(Mar-6,1,May
Since 4.7.10,....are in A.P, a-4, d=3
t,-4H\n-l
.0+{n-1)d=4+(n -1)3 =4+3n-3=3n +1
n

Let S(n)= 14 +
4,7 7.10 (3n-2)(3n+1) 3n+1
Step-l: To prove that S(n)is true for n=l.
1
LHS
(3n-2)(3n+1) (3-2)(3+1) 4
1 1
RHS =
3n +1 3+1 4
:. LHS = RHS, S(1) is true
Step-II: Assume that S(n) is true for n=k.
1 1 ,1 1 k
1.4 4.7 7.10 (3k-2)(3k +1) 3k+1 .()
Step-II: Wehave to prove that S(n) is true for n=k+1
1
(k+1)" term =
(3k+1)(3k+4) Adding on both sides (3k +1)(3k+4) to (1) we get
1,1, 1 1
14 4.7 7.10 (3k-2)(3k +1) (3k +1)(3k +4) 3k+1 (3k +1)(3k +4)
1
k+
L134 +4k+1]
1 1 34 +3k +k+1
3k+1 3k +4 3k +1 3k +4 3k+1 3k +4

3k(k+1)+1(k+1)) (3k+1(k+1) k+1 k+1

(3M
3k +1| 3k +4 (3k +1)(3k +4) 3k+3+1 3(k+1)+1
.S(k+1) is true
.By the principle of finite mathematical induction S(n) is true, Vne N (M)

***4.Show that 2.3+3.4+4.5+.... up to n terms = n(u° +ón +11) VneN

3
(March- 2013, May-200)
Sol: Since 2,3,4 ......e in A.P., a-2.d=1
, =at(7-1)d =2+(n-1)l=n+l
Since 3,4,5,....e in A.P. , a-3, d=1 nAP. t,=at{n-lja
t, =a+(1-1)d=3+(n-1)l=n+2

NARAYANA Page.no
JR-IPE-MATHS-IA
o SOLUTIONS

Let S(n) = +6n +11)

2.3+3.4+4.5 +---+(n+1)(n+2) = 3
(IM)

Step-1: To prove that S(n) is true for n=1.

LHS =(n+1)(n +2) =(1+1)(1+2) =6
RHS nn'+6n +l1) i(1+6 +11)
3
. LHS = RHS, S(1) is true (IM)
Step-ll: Assume that S(n) is true for n=k.
k(k +6k +11) (IM)
Le: 2.3+3.4+4.5+---+{k +1)(k +2) = .(1)
3
Step-ll: We have to prove that S(n) is true for n=k+l
Adding on both sides (k +2)(k +3)to (1) we get
k(* +6k +11) +(k+2)(k +3)
23+3.4+4.5+---+ (k+1)(k +2)+(k +2)(k +3) =
k +6k* +1lk +k'+5k+6= k +6k² +11k +3k²+15k +18 k+9k'+ 26k+18,
3 3
3
+6(k +1)+ 11|
(k+1)(*' +8k +18) (k +1)[*+2k +6k +1 +6+11] (k+1)|(k+i)3
3 3
(3M)
S(k+1) is true (IM)
S(n) is true, tn e N
By the principle of finite mathematical induction
1,1, 1 VneN(Mar-18)
***5. Show that 1.3 3.5 5.7 (2n-1)(2n+1) 2n+1

1 1 1 (IM)
Sol :Let S(n) =13 3.5 5.7 (2n-1)(2n+1) 2n+1
for n=1.
Step-I: To prove that S(n) istrue
1 1 1
LHS =
(2n-1)(2n +1) 1.3 3 .. LHS =RHS, S(1) is true.
(IM)
11
RHS = 3
2n+1 2+1
true for n =k.
Step-II: Assume that S(n) is
1 (IM)
1 1,1 + (1)
(2%-1)(2k+1) 2k+1
i.e; 1.3'3.5 5.7
S(n) is true for n=k+l
Step-II: We have to prove that
get
Adding onboth sides, 2k+)(2k+3) to(1) we
1
(k+1)" term = (2k+1)(2k+3)
Page.no.21
NARAYANA
JR-PE-MATHS-IA

+---+: +1) (2k +1)(2k +3) 2k+1 (2k +\\|2k +3

1.3 3.5 5.7 (2k-1)(2k
|22+3k+1| 2k'+2k+k+1
2k +3 2k +1| 2k +3
2k +3 2k+1|
2k +1
(k+1)(2% +)| k+1
k+1
24(k+1) +1(k +1)|
1
1 2k+3 2k +2+1 2(k+1)+1
2k +3 2k +1|
2k +1
..S(k+l) is true. tne N
principle of finitemathematical induction S(n) is true,
.. By the

***6. Show that

1.2,.3+2.3.4+3.4.5+......up to n terms = n(n+1)(1+2)(n+3)
4 , neN

1.23+23.4+345+---+nn+1)(z+2)="+|2+2)(2+3)
(TS-Mar-2015,17)
Sol: Let S(n) = 4

Step-1: To prove that S(n) is true for n=1.

LHS =n(n+1)(n+2)=1(2)(3) =6
RHS = n(a+)(1+2)(+3)_12)9)(4)
4
=6 (M
:. LHS =RHS S(1) is true.
Step-I1; Assumethat S(n) is true for n=k.

ie; 1.23+2.34+34.5+---+tk(k+1)(k+2)= k(k+1)]|k+2)(k+3) (IM

4
Step-III: We have to prove that S(n) is true for n =k+l
(k+1)" term =(k+1) (k +2) (k +3) Adding on both sides (k +1)(k +2)(k +3), weget
1.2.3+2.34+3.4.5+...k{k+1Xk +2)+(k+1Xk+2Xk+3)=k(k +1Xk+2Xk +3) +(k+1Xk+2Xk+3)
4

-(k=)(k +2)(k+)1k+)(& +2)(k +3)(k+4) 4

. S(k+1) is true
(3M
.Bythe principle offinite mathematical
***7. Prove byMathematical induction S(n) is true, tne N. (M
induction, for all ne N
a+(a+d) +(a+2d)+.... upto nterms
2a+(n-1)d] (Mar-2010,TS-Mar-10)
Sol:Let S(n)=aHa+d)+\a+4)+-aa-)d]-2a+(n-1)d] (IM)

Step-1: To prove that S(n) is true for n=1.

", =a+(n-1)d

NARAYANA
JR-IPE-MATHS-IA SOLUTIONS

LHS =a+(n-1)d=a+(1-)d=a

. LHS =RHS, S() is true (IM)

Step-l: Assumethat S(n) is true forn=k.
(IM)
ic: o+(o+d) +(at 241):+---[o+(k- )d]-[2+(k-)d ... (4)

Step-lll: We have to prove that S(n) is true for n=k+l (IM)

(k+1 tem =(a+kd) Adding on both sides (a +kd) to (1) we get

atla+d)+(a+ 2d) +---+[a+(k-1)d]+a+kd-2a +(k -1)d] +a +kd
2ak +k'd-kd +2a +2kd k'd+ kd +2a(k+1)
2 2

ki(k +)+2a(k+)_(k:pa+kdj- 2a(+1-1)a|

2 2
(2M)
S(k+1) is true. (IM)
N.
. Bythe principle of finitemathematical induction S(n) is true, Vn e
(TS-Mar-20)
***8 Preve by Mathematical induction, for all neN

a+ ar + ar*+ ---Up to nterms =

a(-),1
r-1
(Mar-11,19)

+---+ar= a-1) : n G.P. t. =ar"-| (IM)

Sol :Let S(n)= a+ar + ar r-1

n=1.
Step-I: To prove that S(n) is true for
LHS =ar =ar- =ar

RHS =
a(r-i) ar'-1)
r-1 r-1 (IM)
:. LHS= RHS, S(1) is true.
for n =k.
Step-II: Assumethat S(n) is true
ar'+---+ark- =
a(-1) (IM)
1.e; a+ar + r-1

S(n) is true for n=k+1

Step-II: We have to prove that
both sides ark to (1) we get
(k+1) term = a.r,Adding on
+ar* =
alr-i)
a+ar+ ar +---+ark r-1

as -a+ar"r-ar
-a+ark+! au-i) (3M)
ar-at ar (r-1) r-1
r-l
r-1
Page.no.23
NARAYANA
 SoLUTI0
induction S(n) is true, Vn e N
mathematical n.(May-05,Mar-17
JR-IPE-MATHS-IA offinite
principle 1,20,Ts Mar-A9
.: the
By all positive integers
S(k+1)is true. divisibleby64for
***9. Showthat 49"
+16n-l
+16n-1is
by64
divisible (M
statenent of 49"
Sol:Let S(n) bethe for n=l.
To provethat S{n)is true
Tep-]:
+16(1)-l=65-1=64 (IMy
49" +16n-l= 49
S(1) is true
It is divisible by 64 for n =k.
Step-l;Assumethat S(n) is true integer.
i.e; 49 +16k-1=64m
where mis positive
(IM,
49 =64m -16k+1()
show that Sn) is true for n = k+1
Step-I:We have to
+1)-1= 49*,49 +16k+16-1
Consider, 49t* +16(k
=(64m-16k +1)49 +16k +15(: from())49m-784k +16k+64 =64.49m-768k +64
+16k+15 = 64
= 64m.49-16k.49+49
64(49m-12k +1)=640 where = 49m-12k +1
=64.49m--12,64k +64=
It is divisible by 64.
.. S(k+1) is true.
induction S(n) is true, tn eN (IM)
.:.By the principle of finite mathematical (May -2005)
positive integers n.
***10. Show that 4" -3n-1 divisible by 9 for all
(IM)
Sol: Letthe given statement byp(n)=4"3n-1 0s divisible by 9.
Step-1:-Prove that pn) is true n =1(i.e) prove that p(n) is divisible by9 (IM)
for n = 1

P() =4-3(1)-1
=4-4
=0
Clearly pn) is divisible by9 for n= l
Step-2:-Assume thatp(n) is divisible by 9for n=k(ie.) p(n) is true forn=k. (1M)
P(K) =4 -3k1=9t
’4 =9t +3k +1
Step-3:- Prove that pn) is divisible by 9 for n=k+ 1 (3M)
p(k+1)=4* -3(k+1)-1
=4^4-3k-3-1
=4(9t +3k +1)-3k-3-1
=36t +12k+4-3k -4
= 36t +9k=9(4+k)
Clearly pn) is true for n=k+1
. pn) is divisible by 9for allne N (1M)
***11. Show that 3.52n +23n+1 is divisible by 17 tne N (May-08,10,12)
NARAYANA
) JR-IPE-MATHS-IA SOLUTIONS
Sol:Let S(n)bethe statement of 3.52n+l + 23n+1 is divisible by 17 (IM)
Step-I: To prove that S(n) is true for n=1.
s2n+1 +23n*l =3.5* +2* =3(125)+16=375+16=391=17x23
It is divisible by 17.
.S(1) is true. (IM)
Step-I: Assume that S(n) is true for n=k.
Le: 3.52+ +2** =17m where mis positive integer.
3.52k+ =17m-23k+’(1) (IM)

Step-lIl: We have to show that S(n) is true for n=k+l

Consider, 3.52(*+1)+1 + 23(k+1)+l =3.5k+2+| + 23k+3+1
17m.2525.23k* +8.2k+1
=3.52k+s +23k+1.2 =(17m-23k+1)25 +8.23k+1 =
=17m.25-17.23k+l = 17(25m -21k*!)=170: Where Q= 25m - 2°k*l
(3M)
Itisdivisible by 17. ..S(k+1) is true. (IM)
VneN
-Bythe principle of finitemathermatical induction S(n) is true,
2.42n+1 +33n+1 is divisible by 11.
***12.Use Mathematical induction prove that (1M)
divisible by 11
Sol: Let S(n) be the statement of 2,42n+1 +33n+1 is
n1
Step I: To prove that s(n) is true for
11:
2.42.1+1 +331+1 = 2.4 +3 = 209 =11x19 is divisible by (IM)
..s(n) is true for n=1
Assume that s(n) is ture for n=k
ie 2.42k+! +33k+1 =11m (IM)
2.42k+1 =1lm -33k+1 ..(1) +33k+4 is
n=k+lconsider 2,42(k+l)+l +3k+|) = 2,42k+3
We have to show that s(n) is true for
divisible by11
We have 2.42k+l = 1Lm-33k+l
Multiplying by4?
2.424",4 =(1 1m-3k1).4°
Adding 33k+4 on both sides
2.424*3 +32k** =(1Im-3k*)16+3*
+3h+4 =1 lm.16+3k*. 3 -16|
=llm.16-3k*l,16
=lIm.l6+3k*[11] =11 l6m+3k| (3M)
divisible by |1
..2.42k+3 +33k+4 is
for all n eN (IM)
S(n) is true for n=k+1 mathematical induction s(n) is true
finite
.:. Bythe principle of induction to prove the statement,
***13. Use mathematical
(Mar-15)
|1+-(2+1).
2n+. (TS-Mar-19)
Page.no.25
A
JR-IPE-MATHS-IA
SOLUTL
Sol: Let
Step -1;- for n=1
2n+1)
LH.S=|1+

=(1+3)
=4

RHS=(n+1
-(1+1}
-22 =4
LH.S =R.H.S ----

(IM)
S(1) is true
Step-2:-Assume that s(k) is true forn=k
....) (1M)
(IM)
Step -3:- To prove that s(k+1) is true for n=k+1

(k+1)'
1+
2(k+1)+1|
(k+1y" term
(k+1'
(k+1)' +2k +2+1
(k+1)
(+1+2k+2k +3)
(k+1)'
k' +4k+4
(k+1'
(k+2)
(k+1)]'
Multipying (k+1)" term inequation (1)

(2M)
(k+1
-(k+2'
s(k+1) is true
. bvthe priniciple of tnite
mathematical induction s(n) is true neN (IM)
NARAYANA
Page.no.
JR-IPE-MATHS-IA
SOLUTIONS
**14. Use mathematical
induction to prove the statement,
k=l
n(2n-1)(2n+1) for all ne N
3

Sol:
k=l
n(2n-1)(2n+1)
3

Let s(n)=1l'+3 +s' +7....(2n-1' "(2-1)(2n+1) ----(IM)

step (1): for n= 1
L.H.S= (2n-1)'
(2-1)
=1'
=1

RH.S= n(2n-1)(2n+1)
3

I(2-)(2+1)
3
1.1.3
3
3

=1
L.H.S=R.H.S
(IM)
S(1) is true
Step (2):Assume that s(k) is true for n=k
s(k) =1 +3² +s +7°....+(2k -1 =
k(2k-1)(2k+1)
3 2.) ----(IM)

=k+1
Step (3) :To prove that s(k+1) is true for n

P+3 +5² +7'..(2k

+1)2_(k+1)(2(k+1)-1)|2(k+1)+1)
3

=S(k)+(2k +1)'
k(2k -1)(2k +)(2k+1)
3

k(2%-1)(2k +1)+3(2k +1)

3

(2k +1)(&(2k-1)+3(2k +1)

3
(2k +1)(2k²-k+6k +3)

Page.no.27
NARAYANA
 SOLUTI0
JR-IPE-MATHS-IA

(2k +1)(2k'+5k+3)
3k +3)
(2k +1)(2A' +2k +
+)
(2k+1)(2A(k+1)+3(k
3

(2k +1)(k +1)(2k +3)

(k+1)(2(% +I)-1)(2(k+1)+1) ---- (3M)

3
(n) istrue neN ----(M
s(h+1) is true mathematicalinduction s
conclusion: by the priniciple
offinite statement,
induction to prove the
*15. Use mathematical
+..... upto nterms= 16n²(n+1)'.
4' +8'+12
upto nterms =16n (n+1)
Sol: 4'+8° +12+....
4,8,12.....are in A.P
n" term , =a+{7-1)d
=4+ (n -1)4
=4+ 4n -4
= 4n

n" term of thegiven series =(4n) ----(1M)

+....41n)=16n'(n+1)'
Let s(n) =4' +8 +12'
step (1):- for n=1
LH.S=(4n)'
=(4.1)
=4
=64
RH.S= 16n' (n+1)'
=16.I'(1+1)'
= J6.22
=16.4
---- (IM)
=64
LH.S=R.H.S
s(l) is true
n=k --- (|M)
step (2):-Assume that s(k) is true for
=4' +8' +12' +.... .(4%)' =\6k'(k +1)
n=k+|
step (3) :-To prove that s(k+ 1)is true for
4+8 +12'+... 4(k+1))=l6(k+1)'(k+2)'
Page.no.
NARAYANA
 JR-IPE-MATHS-IA
SQLUTIONS
s(k)+64(k+1)
=16*° (k +1) +64(k +1)'
-16(k +1) (* +4(k +)
=16(k+1)"(*+4k +4)
-16(k +1)' (k +2)'
s(k+l) is true
---- (3M)
conclusion:by the priniciple of finite mathematical induction s(n) is true y neN ---(M)
s6 Use Mathematical lnduction to prove the statement
2+3.2 +4.2* +--- upto nterms = n.2" ne N (May-07)
Sol: Since 2,3,4,.... are in A.P., a-2, d=1
L,=at{7-l)d =2+(n--1)1=n+1
Since 1,2.2 ---- are in GP.,a=l,2

Let S(n)= 2+3.2 +4.2*+---+(n+1)2* =n.2" (IM)

Step-I:- To prove that S(n) is true for n=1.

LHS =(n+1)2r =(1+1)2 =2
RHS = n.2" = 1.2' =2
.LHS =RHS, S(1) is true. (IM)
Step-IL:-Assume that S(n) is true for n =k
(IM)
ie; 2+3.2-+4.2 +---+(k+1)2= k.2. ()

Step-III:-We have to show that S(n) is true for n=k+1

(k+2)2*O() we get
(k+1)M term = (k+2)2*, Adding on bothsides
2+3.2+4.2 + ---+(k +1)2* +(k+2)2 = k.2 +(k+2)2*
=(k+1)2*
- (k+k+2)2 =(2k +2)2 =2(k+1)2' (3M)
:.S(k+1)is true (IM)
mathematical induction S(n) is true, tn e N
.:. Bythe principle of finite
Showthat y" + y" is divisible by x+y, ifm is an odd
*17. () Using Mathematical Induction,
natural numbers.
natural number and x, y are
Sol: Since m is an odd natural number.
Let m-2nt1 (IM)
S(n) be the statement of x2n+ + yn* is divisible by x +y
Let
true for n1.
Step-I:To prove that S(n) is
=(x+y)lx-y+y)
'+yl =+yn =x+y' (IM)
ltis divisible by x +y..S(1) is true.
Page.ne.29
NARAYANA
 yand
ofx
*'+y**(x*-y'
p"y'=Px-(*-y)y24:!
[(r+y)p-yt
)
*"x+y"y'that
Show
eN Induction,
.x-yy eN
terms y2k+2+1 Vn y
and
n=k+1 (1+y)p.x-=
whereQ
true,
xterms -(*-y)l(mx+
y) tn
true,
inpolynomial is Mathematicalx-y n=k+1 y'
+t2+1 S(n)
in
mx+ is
S(n)
for induction
y)yt]=(*+y)Q by polynomial for =(r-y)
n=k. true =x2k = y)y* n=1.
divisible
true
induction
a -() is y(k++1)+! +y*".,y² using
mathenmatical for
=k.
is
for iPs S(n) n }-'y
true where (1+y)(- y is true
YneN for is S(n)
.....(1)
=sx-y.y-(r-y)max
+(*-y)y' mathematical
that * y-y true m
where that Consider,
yhlye
=(1+r)p-".
isS(n) r)p. show r'-y'hr x
and
is
S(n)
+y*-(t 2(+1)+1 (*-
x-y,
finitenumbers of x-y"
=x-y'
=x-y
is
S(n)
=(x-y)m
show -[(<-y)m+
that to
have )[P'-of by statement
y)pr'- that
that +y.
x=(x-y)m
to (x-y) finite
=(u+y)p principle divisible prove
t
:-Assume
JR-IPE-MATHS-IA true. natural Step-II:-Assumehave of
:-We the
x
by by true principle
(r+ =(*+ We
Step-IIlConsider is To true
.:S(1)
is
divisible divisible
,4l
1.e:
Step-II ,
the are is beS(n) Step-I:-
S(k+1) x-y :-
Step-III
is
S(k+1)
= By x,y x-" Bvthe
.. .. If :Let is
It ie; Itis
() Sol :. .