How do you find the value of a if the infinite geometric series a+ a^2 + 2^3 + ...... =4a, a cannot equal to 0?

1 Answer
George C.
Nov 27, 2015

Assuming the 2^3 should have been a^3...

(1-a)4a = (1-a)(sum_(n=1)^oo a^n) = a

Hence a = 3/4

Explanation:

(1-a)(sum_(n=1)^N a^n)

=sum_(n=1)^N a^n - a sum_(n=1)^N a^n

=a+color(red)(cancel(color(black)(sum_(n=2)^N a^n)))-color(red)(cancel(color(black)(sum_(n=2)^N a^n)))-a^(N+1)

=a-a^(N+1)

If abs(a) < 1 then

(1-a)(sum_(n=1)^oo a^n) = lim_(N->oo) (a-a^(N+1)) = a

(If abs(a) >= 1 then the sum does not converge)

So:

(1-a)4a = (1-a)(sum_(n=1)^oo a^n) = a

We are told that a!=0, so we can divide both sides by a to get:

4(1-a) = 1

So: 1-a = 1/4

So: a = 1-1/4 = 3/4

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