Question

Let $S$ be the sum , $P$ the product and $R$ the sum of reciprocals of $n$ terms in a $G.P$. Prove that $P^2{R}^n=S^n$.

Answer 1

Let the G.P. be $a,ar,{ar}^2,{ar}^3,...,{ar}^{n-1},...$
According to the given information,
$S=\frac{a(r^n-1)}{r-1}P=a^n\times r^{1+2+...+4n-1}=a^n{r}^{\frac{n(n-1)}{2}}R=\frac{1}{a}+\frac{1}{ar}+...+\frac{1}{{ar}^{n-1}}=\frac{r^{n-1}+r^{n-2}+...+r+1}{{ar}^{n-1}}=\frac{1(r^n-1)}{(r-1)}\times \frac{1}{{ar}^{n-1}}=\frac{r^n-1}{{ar}^{n-1}(r-1)}\therefore P^2{R}^n=a^{2n}{r}^{n(n-1)}\frac{(r^n-1{)}^n}{a^n{r}^{n(n-1)}{(r-1)}^n}=\frac{a^n(r^n-1{)}^n}{{(r-1)}^n}={\left[\frac{a(r^n-1)}{(r-1)}\right]}^n=S^n$
Hence, $P^2\cdot R^n=S^n$