Question

Assertion :For $n\ge 1$, let $C_r=\left(\begin{array}{c}n\\ r\end{array}\right)$, $0\le r\le n$

${lim}_{x\to \infty }{\sum }_{r=0}^n\frac{C_r}{(r+3)n^r}=e-2$

Reason: ${\sum }_{r=0}^n\frac{C_r}{(r+3)n^r}={\int }_0^1{(1+\frac{x}{n})}^n{x}^{2}dx$

• A. Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
• B. Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
• C. Assertion is correct but Reason is incorrect
• D. Assertion is incorrect but Reason is correct

Answer 1

The correct option is A Both Assertion and Reason are correct and Reason is the correct explanation for Assertion

$\sum _{r=0}^n{C}_r{t}^r={(1+t)}^n\Rightarrow \sum _{r=0}^n{C}_r{t}^{r+2}=t^2{(1+t)}^n$
$\Rightarrow {\int }_0^x\left(\sum _{r=0}^n{C}_r{t}^{r+2}\right)={\int }_0^x{t}^2{(1+t)}^{n}dt\Rightarrow \sum _{r=0}^n\frac{C_r}{r+3}{x}^{r+3}={\int }_0^x{t}^2{(1+t)}^{n}dt$
Putting $x=\frac{1}{n}$
$\sum _{r=0}^n\frac{C_r}{r+3}{\left(\frac{1}{n}\right)}^{r+3}={\int }_0^x{t}^2{(1+t)}^{n}dt\Rightarrow \sum _{r=0}^n\frac{C_r}{(r+3)n^r}=n^3{\int }_0^{\frac{1}{n}}{t}^2{(1+t)}^{n}dt$
Putting $t=\frac{u}{n}$ to obtain
$\sum _{r=0}^n\frac{C_r}{(r+3)n^3}={\int }_0^1{(1+\frac{u}{n})}^n{u}^{2}du$
Taking limit $n\to \infty$, we get
$\underset{n\to \infty }{lim}\sum _{r=0}^n\frac{C_r}{(r+3)n^r}={\int }_0^1\underset{n\to \infty }{lim}{(1+\frac{u}{n})}^n{u}^{2}du={\int }_0^1{u}^2{e}^{u}du=e^u(u^2-2u+2){]}_0^1=e-2$