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

Nov 27, 2015

Assuming the ${2}^{3}$ should have been ${a}^{3}$...

$\left(1 - a\right) 4 a = \left(1 - a\right) \left({\sum}_{n = 1}^{\infty} {a}^{n}\right) = a$

Hence $a = \frac{3}{4}$

#### Explanation:

$\left(1 - a\right) \left({\sum}_{n = 1}^{N} {a}^{n}\right)$

$= {\sum}_{n = 1}^{N} {a}^{n} - a {\sum}_{n = 1}^{N} {a}^{n}$

$= a + \textcolor{red}{\cancel{\textcolor{b l a c k}{{\sum}_{n = 2}^{N} {a}^{n}}}} - \textcolor{red}{\cancel{\textcolor{b l a c k}{{\sum}_{n = 2}^{N} {a}^{n}}}} - {a}^{N + 1}$

$= a - {a}^{N + 1}$

If $\left\mid a \right\mid < 1$ then

$\left(1 - a\right) \left({\sum}_{n = 1}^{\infty} {a}^{n}\right) = {\lim}_{N \to \infty} \left(a - {a}^{N + 1}\right) = a$

(If $\left\mid a \right\mid \ge 1$ then the sum does not converge)

So:

$\left(1 - a\right) 4 a = \left(1 - a\right) \left({\sum}_{n = 1}^{\infty} {a}^{n}\right) = a$

We are told that $a \ne 0$, so we can divide both sides by $a$ to get:

$4 \left(1 - a\right) = 1$

So: $1 - a = \frac{1}{4}$

So: $a = 1 - \frac{1}{4} = \frac{3}{4}$