Question

The first term of a certain infinitely decreasing geometric progression is $I$ and its sum is $S$. Find the sum of the geometric progression which is formed by the squares of the terms of the initial progression.

Answer 1

$S=I+Ir+I{r}^2....\infty =\frac{I}{1-r}$

New series$=I^2+I^2{r}^2+I^2{r}^4+I^2{r}^6.....\infty$

Sum $=\frac{I^2}{1-r^2}$

$S=\frac{I}{1-r},1-r=\frac{I}{S},r=1-\frac{I}{S}$

Sum$=\frac{I}{1-r}\times \frac{I}{1+r}\Rightarrow S\times \frac{I}{1+1-\frac{I}{S}}$

$\Rightarrow \frac{SI}{2S-I}\times S\Rightarrow \frac{S^{2}I}{2S-I}$