Question

Find the sum to infinity of the following arithmetic – geometric sequence 1, 2/4, 3/16, 4/64, …​

Answer 1

Answer:

The answer is $\frac{16}{9}$

Step-by-step explanation:

Let the sum be $x$

$\Rightarrow x=1+\frac{2}{4}+\frac{3}{16}+\frac{4}{64}+...\rightarrow (1)$

Multiply both sides by $\frac{1}{4}$ (common ratio)

$\Rightarrow \frac{1}{4}\times x=\frac{1}{4}+\frac{2}{16}+\frac{3}{64}+\frac{4}{256}+...\rightarrow (2)$

Now subtract (2) from (1)

$\Rightarrow x-\frac{x}{4}=1+(\frac{2}{4}-\frac{1}{4})+(\frac{3}{16}-\frac{2}{16})+...\\\Rightarrow \frac{3x}{4}=1+\frac{1}{4}+\frac{1}{16}+...\\\Rightarrow \frac{3x}{4}=\frac{1}{1-\frac{1}{4}}=\frac{4}{3}\\\therefore x=\frac{16}{9}$

( Using $S_\infty =\frac{a}{1-r}$ )