Question

${2.}^n{C}_0+2^2.\frac{{}^n{C}_1}{2}+2^3.\frac{{}^n{C}_2}{3}+\dots +2^{n+1}.\frac{{}^n{C}_n}{n+1}=$

• A. $\frac{3^{n+1}-1}{2(n+1)}$
• B. $\frac{3^{n+1}-1}{n+1}$
• C. $\frac{3^n-1}{n+1}$
• D. $\frac{3^n+1}{n+1}$

Answer 1

The correct option is B $\frac{3^{n+1}-1}{n+1}$

We have,
$(1+x{)}^n{=}^n{C}_0{+}^n{C}_1(x\left){+}^n{C}_2\right(x{)}^2+...{+}^n{C}_{n-1}\left(x{)}^{n-1}{+}^n{C}_n\right(x{)}^n$
Integrating wrt. $x$,
$\frac{(1+x{)}^{n+1}}{n+1}{=}^n{C}_0\left(x\right)+\frac{{}^n{C}_1(x{)}^2}{2}+\frac{{}^n{C}_2(x{)}^3}{3}+...+\frac{{}^n{C}_n(x{)}^{n+1}}{n+1}+k$
Substitute $x=0$
$\frac{1}{n+1}=k$
Substitute $x=2$,
$\frac{3^{n+1}}{n+1}{=}^n{C}_0\left(2\right)+\frac{{}^n{C}_1(2{)}^2}{2}+\frac{{}^n{C}_2(2{)}^3}{3}+...+\frac{{}^n{C}_n(2{)}^{n+1}}{n+1}+\frac{1}{n+1}$
Hence,
${}^n{C}_0\left(2\right)+\frac{{}^n{C}_1(2{)}^2}{2}+\frac{{}^n{C}_2(2{)}^3}{3}+...+\frac{{}^n{C}_n(2{)}^{n+1}}{n+1}=\frac{3^{n+1}-1}{n+1}$
Hence, option B is the correct answer.