Question

$3C_0+\frac{3^2}{2}{C}_1+\frac{3^3}{3}{C}_2...upto(n+1)term$ equals

• A. $\frac{1}{n+1}(4^{n+1}-1)$
• B. $\frac{n}{n+3}(4^{n+1}+1)$
• C. $\frac{1}{n+2}(4^{n+1}-1)$
• D. $\frac{1}{n+1}(4^{n+1}+1)$

Answer 1

Answer: (A)

$\frac{1}{n+1}(4^{n+1}-1)$

Consider
$(1+x{)}^n={}^n{C}_0+{}^n{C}_{1}x+{}^n{C}_2{x}^2+...{}^n{C}_n{x}^n$
Integrating both sides with respect to x.
$\frac{(1+x{)}^{n+1}}{n+1}+C=x{}^n{C}_0+\frac{x^2}{2}{}^n{C}_1+\frac{x^3}{3}{}^n{C}_2+...{}^n{C}_n\frac{x^{n+1}}{n+1}$
At $x=0$ RHS=0.
Hence
$C=\frac{-1}{n+1}$
Thus the overall expression is
$\frac{(1+x{)}^{n+1}-1}{n+1}=x{}^n{C}_0+\frac{x^2}{2}{}^n{C}_1+\frac{x^3}{3}{}^n{C}_2+...{}^n{C}_n\frac{x^{n+1}}{n+1}$
Let $x=3$
Hence
$\frac{4^{n+1}-1}{n+1}=3{}^n{C}_0+\frac{3^2}{2}{}^n{C}_1+\frac{3^3}{3}{}^n{C}_2+...{}^n{C}_n\frac{3^{n+1}}{n+1}$
Hence the above series can be written as
$=\frac{4^{n+1}-1}{n+1}$