Question

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

Answer 1

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`