Question

If $\sum _{r=0}^n{\left(\frac{r+2}{r+1}\right)}^n{C}_r=\frac{2^8-1}{6}$, then $n$ is

• A. $8$
• B. $4$
• C. $6$
• D. $5$

Answer 1

The correct option is D $5$

Expanding we get
$2+\frac{3}{2}{}^n{C}_1+\frac{4}{2}{}^n{C}_2+....\frac{n+2}{n+1}{}^n{C}_n$
Consider
$(1+x{)}^n=1+x^1{}^n{C}_1+x^2{}^n{C}_2+x^3{}^n{C}_3+...+x^n{}^n{C}_n$
Integrating with respect to x we get
$\frac{(1+x{)}^{n+1}}{n+1}+c=x+\frac{x^2}{2}{}^n{C}_1+\frac{x^3}{3}{}^n{C}_3+...\frac{x^{n+1}}{n+1}{}^n{C}_n$
Here $C=\frac{-1}{n+1}$ hence we get
$\frac{(1+x{)}^{n+1}-1}{n+1}=x+\frac{x^2}{2}{}^n{C}_1+\frac{x^3}{3}{}^n{C}_3+...\frac{x^{n+1}}{n+1}{}^n{C}_n$
Multiplying with x, we get
$\frac{x(1+x{)}^{n+1}-x}{n+1}=x^2+\frac{x^3}{2}{}^n{C}_1+\frac{x^4}{3}{}^n{C}_3+...\frac{x^{n+2}}{n+1}{}^n{C}_n$
Now differentiating with respect to x, we get
$\frac{1}{n+1}\left[\right(1+x{)}^{n+1}+x(n+1)(1+x{)}^n-1]=2x+\frac{3x^2}{2}{}^n{C}_1+\frac{4x^3}{3}{}^n{C}_3+...\frac{(n+2)x^{n+1}}{n+1}{}^n{C}_n$
Substituting x=1, we get
$2+\frac{3}{2}{}^n{C}_1+\frac{4}{2}{}^n{C}_2+....\frac{n+2}{n+1}{}^n{C}_n$
$=\frac{1}{n+1}\left[\right(2{)}^{n+1}+(n+1)(2{)}^n-1]$
$=\frac{1}{n+1}\left[2^n\right(n+3)-1]$
Comparing the coefficients with the given question we get
$n+1=6$
$n=5$