Understanding Engineering Mathematics

because all terms but the first and last cancel out.
So

Sn=

a( 1 −rn)

1 −r

Example

For the series

∑^5

r= 1

1

3 r−^1

= 1 +

1

3

+

1

9

+

1

27

+

1

81

a= 1 ,r=

1

3

,S 5 =

1

(

1 −

(

1

3

) 5 )

1 −

1

3

=

121

81

You can practice your skills in handling fractions by checking this directly.
Another interesting series is thearithmetic series

S=a+(a+d)+(a+ 2 d)...

where each successive term is formed by adding a ‘common difference’d, so that thenth
term isa+(n− 1 )d. In this case we have

Sn=a+(a+d)+(a+ 2 d)+···+(a+(n− 1 )d)

and a neat way to sum this is to reverse the series and add corresponding terms of the two
results:

Sn=a+(a+d)+(a+ 2 d)+···+(a+(n− 1 )d)

=(a+(n− 1 )d)+(a+(n− 2 )d)+···+a

So

2 Sn=( 2 a+(n− 1 )d)+( 2 a+(n− 1 )d)+···+( 2 a+(n− 1 )d)

=n( 2 a+(n− 1 )d)

and therefore

Sn=^12 n( 2 a+(n− 1 )d)

for the arithmetic series.

Solution to review question 3.1.9

The sum tonterms of the finite geometric series with common ratior,

first termais

Sn=

a( 1 −rn)

1 −r