Question

Let $A_r$ be the area bounded by the curve $y=x^r(r\ge 1)$ and the line $x=0,y=0$ and $x=\frac{1}{2}.$ If $\sum _{r=1}^n\frac{2^r{A}_r}{r}=\frac{1}{3},$ then the value of $n$ is

Answer 1

$A_r=\underset{0}{\overset{1/2}{\int }}{x}^{r}dx$
$\Rightarrow A_r=\frac{1}{2^{r+1}(r+1)}$
$\Rightarrow 2^r{A}_r=\frac{1}{2(r+1)}\Rightarrow \frac{2^r{A}_r}{r}=\frac{1}{2(r+1)r}$
Hence, $\sum _{r=1}^n\frac{2^r{A}_r}{r}=\frac{1}{2}\sum _{r=1}^n(\frac{1}{r}-\frac{1}{r+1})$
$=\frac{1}{2}(1-\frac{1}{n+1})$

Given, $\frac{1}{2}(1-\frac{1}{n+1})=\frac{1}{3}$
$\Rightarrow 1-\frac{2}{3}=\frac{1}{n+1}$
$\Rightarrow \frac{1}{3}=\frac{1}{n+1}\Rightarrow n=2$