Seven days to go for JEE 2016 - 1

Refer to Definite Integration

$S_k=\displaystyle\int_0^1x^2(1-x)^k\mathrm{d}x=\displaystyle\int_0^1(1-x)^2x^k\mathrm{d}x$ $\color{teal}{\small{(\text{Use } \displaystyle\int_a^bf(x)\mathrm{d}x= \displaystyle\int_a^bf(a+b-x)\mathrm{d}x)}}$ Hence,

$\large{\begin{aligned}&\displaystyle\sum_{k=1}^{\infty}\displaystyle\int_0^1(1-x)^2x^k\mathrm{d}x\\&=\displaystyle\int_0^1(1-x)^2\underbrace{\displaystyle\sum_{k=1}^{\infty}x^k}_{\color{#20A900}{\text{Infinite GP}}}\mathrm{d}x\\&=\displaystyle\int_0^1(1-x)^{\cancel{2}}\left(\dfrac{x}{\cancel{1-x}}\right)\mathrm{d}x\\&=\displaystyle\int_0^1(x-x^2)\mathrm{d}x\\&=1/2-1/3=1/6\large\end{aligned}}$

$\Huge\therefore 1+6=\color{#D61F06}{\boxed{\color{#0C6AC7}{7}}}$

By inspection: $\displaystyle { S }_{ k }=\int _{ 0 }^{ 1 }{ { x }^{ 2 }{ \left( 1-x \right) }^{ k }dx } =B\left( 3,k+1 \right)$

Here, $B(x,y)$ is beta function .

Now we have: $\sum _{ k=1 }^{ \infty }{ { S }_{ k } } =\sum _{ k=1 }^{ \infty }{ \frac { 2\Gamma \left( k+1 \right) }{ \Gamma \left( k+4 \right) } }$

On splitting the terms using partial fractions, we get: $\sum _{ k=1 }^{ \infty }{ { S }_{ k } } =\sum _{ k=1 }^{ \infty }{ \left[ \frac { 1 }{ 2\left( k+3 \right) } +\frac { 1 }{ 2\left( k+1 \right) } -\frac { 1 }{ k+2 } \right] }$

This, clearly, is a telescoping series. This telescopes to: $\displaystyle \sum _{ k=1 }^{ \infty }{ { S }_{ k } } =\frac { 1 }{ 6 }$ .

Here, $p=1$ and $q=6$ . Hence, $\boxed{p+q=7}$