Sigma notation

Sigma notation

 = Sum

n

 General term
▪ Where n is the number of term in the sequence
▪ k indicates from which term we must count from
k ▪ General term is either an AS or a GS
 Examples of the notation

Please note all of these examples start counting from Term 1

(k = 1, i = 1 and r = 1)
 Example 1

10
n
 ( 2k − 6 ) S n =  2a + ( n − 1) d 
k =1 2
T1 = 2 (1) − 6 = −4 10
S10 =  2 ( −4 ) + (10 − 1)( 2 ) 
T2 = 2 ( 2 ) − 6 = −2 2
T3 = 2 ( 3) − 6 = 0 S10 = 5  2 ( −4 ) + (10 − 1)( 2 ) 
d = T2 − T1 = T3 − T2 S10 = 5  −8 + 18 = 50
−2 − (−4) = 0 − ( −2)
2 = 2  It is an Arithmetics sequence
 8

 5 ( −2 )
i

i =1

T1 = 5 ( −2 ) = −10
1

T2 = 5 ( −2 ) = 20
2 a ( r n − 1)
Sn =
T3 = 5 ( −2 ) = −40
3 r −1
T2 T3
r= =
T1 T2 S8 =
(
−10 ( −2 ) − 1
8
)
−2 − 1
20 −40
=
−10 20 −10 ( 255 )
S8 =
−2 = −2 −3
 It is a Geometric sequence S8 = 850
 

 2 ( 4)
1− r

r =1

T1 = 2 ( 4 )
1−1

T1 = 2 ( 4 ) = 2
0

T2 = 2 ( 4 )
1− 2
a
S =
T2 = 2 ( 4 )
−1
=
1 1− r
2 2 8 2
T3 = 2 ( 4 )
1−3
S = = =2
1 3 3
1 1−
T3 = 2 ( 4 )
−2
= 4
8
1
r=
4
 More complex examples

Please note all of these examples start counting

from different terms:
(k = 3, k = -2 and i = 5)
 The sequence has got 10 terms but we must start counting
from term 3:

 n = 10 − 3 + 1 = 8
n
10
S n =  2a + ( n − 1) d 
 ( 5k + 4 )
k =3
2
8
T1 = 5k + 4 = 5 ( 3) + 4 = 19 S8 =  2 (19 ) + ( 8 − 1)( 5 ) 
2
T2 = 5k + 4 = 5 ( 4 ) + 4 = 24 S8 = 4 38 + 35
T3 = 5k + 4 = 5 ( 5 ) + 4 = 29 S8 = 4  73 = 292
a = 19, d = 5
 The sequence has got 6 terms but we must start counting
from term -2:

 n = 6 − ( −2 ) + 1 = 9
6

 2 ( 3)
k +2

k =−2
a ( r n − 1)
T1 = 2 ( 3) = 2 ( 3) = 2
−2 + 2
Sn =
0

T2 = 2 ( 3)
−1+ 2
= 2 ( 3) = 6
1 r −1
T3 = 2 ( 3)
0+ 2
= 2 ( 3) = 18
2

S8 =
(
2 ( 3) − 1
9
)
T2 T3
r= = 3 −1
T1 T2
6 18 2 (19682 )
= S8 =
2 6 2
3=3
S8 = 19682
 It is a Geometric sequence
 The sequence has infinite number terms but we must
start counting from term 5. In this example we need only
find a and r as it is the sum to infinity.


 ( 3)
6 −i

i =5

T1 = ( 3) = ( 3) = ( 3) = 3
6 −i 6 −5 1
a
S =
T2 = ( 3)
6 −i
= ( 3)
6−6
= ( 3) = 1
0 1− r
3 9 1
1 S = = =4
T3 = ( 3) = ( 3) = ( 3)
6 −i 6−7 −1
= 1 2 2
3 1−
3
1
r=
3
Writing series in sigma notation
Writing series in sigma notation
Classwork / homework

Exercise 7 Page 23 & 24

 a) 1; 2; 4; 7; 10 and 12
 c) 1; 3; 5 and 7
Finding the number
of terms in Sigma
notation (n)
 m n
S n =  2a + ( n − 1) d 
 ( k + 1) = 1595
k =1
2
m
T1 = 1 + 1 = 2 1595 =  2 ( 2 ) + ( m − 1)(1) 
2
T2 = 2 + 1 = 3 3190 = m  4 + m − 1
T3 = 3 + 1 = 4 3190 = m  m + 3
a = 2, d = 1 3190 = m 2 + 3m
m 2 + 3m − 3190 = 0
( m − 58 )( m + 55 ) = 0
m − 58 = 0 or m + 55 = 0
m = 58 m  −55
 n
S n =  2a + ( n − 1) d 
m

 ( 2 − 3 p ) = −7085
p =5
2
n
−7085 =  2 ( −13) + ( n − 1)( −3) 
T1 = 2 − 3(5) = −13 2
−14170 = n  −26 − 3n + 3
T2 = 2 − 3(6) = −16
−14170 = n  −23 − 3n 
T3 = 2 − 3(7) = −19
−14170 = −23n − 3n 2
3n 2 + 23n − 14170 = 0
a = −13, d = −3 ( n − 65 )( 3n + 218 ) = 0
n = m − 5 +1
n = 65 or n  −
218 65 = m − 5 + 1
3
m = 69
 m
1 k −1 1640

k =−2 2
( 3) =
27 Sn =
a ( r n − 1)
r −1
1 n
1640 54 (
1 −2−1 1 −3 1 3 − 1)
T1 = ( 3) = ( 3) =
2 2 54 =
27 3 −1
1 −1−1 1 −2 1
T2 = ( 3) = ( 3) = 1 n n = m − ( −2 ) + 1
1640 54 (
2 2 18 3 − 1)
1 0−1 1 −1 1
T3 = ( 3) = ( 3) = 27
=
2
8 = m + 2 +1
2 2 6
1
3280 1 n
= ( 3 − 1) m=5
a = ,r = 3 27 54
54 6560 = 3n − 1
6561 = 3n
38 = 3n
n=8