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