Question

If $P_n$ is the sum of a $GP$ upto $n$ terms. Show that $(1-r)\frac{d{p}_n}{dr}=n.P_{n-1}-(n-1)P_n$.

Answer 1

$P_n=$ Sum of n terms in G.P.

$P_n=\frac{a(r^n-1)}{r-1}$, a is a first term,r is a common ratio.

$\frac{d{P}_n}{dr}=\frac{an{r}^{n-1}(r-1)-a(r^n-1)}{(r-1{)}^2}\frac{d{P}_n}{dr}=\frac{an{r}^n-an{r}^{n-1}-a{r}^n+a}{(r-1{)}^2}\frac{d{P}_n}{dr}=\frac{a{r}^n(n-1)-an{r}^{n-1}+a}{(r-1{)}^2}\frac{d{P}_n}{dr}=\frac{(n-1)a{r}^n-an+a+an-an{r}^{n-1}}{(r-1{)}^2}\frac{d{P}_n}{dr}=\frac{(n-1)a(r^n-1)-\left(n\right)\left(a\right)(r^{n-1}-1)}{(r-1{)}^2}\frac{d{P}_n}{dr}=\frac{(n-1)\left[\frac{a\left(r^{n-1}\right)}{r-1}\right]-n\left[\frac{a\left(r^{n-1}\right)}{r-1}\right]}{(r-1)}(r-1)\frac{d{P}_n}{dr}=(n-1)P_n-n{P}_{n-1}(1-r)\frac{d{P}_n}{dr}=n{P}_{n-1}-(n-1)P_n$