Question

(a) Let a denote the first term and r the common ratio of an exponential sequence. (i) show that the general rule for finding the sum, Sn' of the first n term of the sequence is give by (a(1-r^n))/(1-r),r≠1. (ii) Hence deduce the sum of infinity of an exponential sequence with |r|<1.

Answer 1

Since \[
|r|<1
\]We have \[
n\to \infty \implies r^n\to0
\]

Answer 2

is that all

Answer 3

You still must do part (i)

Answer 4

could you please solve

Answer 5

\[
(1-r)\sum_{k=0}^{n-1} r^k = \sum_{k=0}^{n-1} r^k -r^{k+1} = 1-r^n
\]So \[
\sum_{k=0}^{n-1} r^k=\frac{1-r^n}{1-r}
\]

Answer 6

You make it a telescoping sum

Answer 7

what do you imply by telescoping sum

Answer 8

For the terms \(2,3,\dots , n-1\) they cancel each other out becaues you have \(r^0-r^1+r^1-r^2+r^2-\dots\)

Answer 9

What next?