Question

If$C_0,C_1,C_2..,C_n$ denote the binomial coefficients in the expansion of

${(1+x)}^n,then{C}_0+(C_1/2)+(C_2/3)+\dots ..C_n/(n+1)=$

• A. $[2^{n+1}–1]/(n+1)$
• B. $[2^n–1]/(n+1)$
• C. $[2^{n-1}–1]/(n-1)$
• D. $[2^{n+1}–1]/(n+2)$

Answer 1

The correct option is A

$[2^{n+1}–1]/(n+1)$

Explanation for the correct options:

Finding the value of $c_0+\left(\raisebox{1ex}{$c_1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right)+\left(\raisebox{1ex}{$c_2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.\right)+....\raisebox{1ex}{$c_n$}\!\left/ \!\raisebox{-1ex}{$\left(n+1\right)=$}\right.$

$$\begin{gathered} {\left(1+x\right)}^n=1+{}^{n}c_{1}x+{}^{n}c_2{x}^2+{}^{n}c_n{x}^n \\ {\int }_0^1{\left(1+x\right)}^{n}dx=1+\frac{{}^{n}c_1}{2}+\frac{{}^{n}c_3}{3}+\frac{{}^{n}c_4}{2}...+\frac{{}^{n}c_n}{n+1} \end{gathered}$$

Hence

$$\begin{gathered} {\int }_0^1{(1+x)}^{n}dx={{\left[\frac{{\left(1+x\right)}^{n+1}}{n+1}\right]}_0}^1 \\ \mathbf{=}\frac{{\mathbf{2}}^{\mathbf{n}\mathbf{+}\mathbf{1}}\mathbf{-}\mathbf{1}}{\mathbf{n}\mathbf{+}\mathbf{1}} \end{gathered}$$

Hence option(A) is correct.