Question

Let $S$ be the sum, $P$ be the product and $R$ is the sum of reciprocals of $n$ terms in a GP.

Prove that $P^2{R}^n=S^n$.

Answer 1

Let G.P. be $a,ar,a{r}^2,....a{r}^{n-1}$

$S=\frac{a(r^n-1)}{r-1};P{a}^n{r}^{1+2+...+n-1}=a^n{r}^{\frac{(n-1)n}{2}}$

$R=\frac{1}{a}+\frac{1}{a^r}+\frac{1}{a{r}^2}=\frac{1}{a}(r^0+r^{-1}+r^{-2}+....)$

$=\frac{1}{a}\times r^0\left(\frac{r^n-1}{r^n-1}\right)=\frac{1}{a}\left(\frac{1-r^n\times r}{r^n(1-r)}\right)=\frac{(1-r^n)\left(r^{1-n}\right)}{a(1-r)}$

$\Rightarrow R=\frac{r^{1n}}{a}\left(\frac{r^n-1}{r-1}\right)$

$=P^2{R}^n=a^{2n}{r}^{n(n-1)}\times \frac{r^{n-n^2}}{a^n}{\left(\frac{r^n-1}{r-1}\right)}^n=a^n{\left(\frac{r^n-1}{r-1}\right)}^n=S^n$