Question

Sum of the series
$S={\sum }_{r=0}^n\frac{3^{r+4}\left({{}^{n}C}_r\right)}{{{}^{r+4}C}_4}+{\sum }_{r=0}^3\frac{{{}^{n+4}C}_r{3}^r}{{{}^{n+4}C}_4}$ is

• A. $4^{n+4}$
• B. $4^{n+4}\left({{}^{2n}C}_4\right)$
• C. $\frac{4^{n+4}}{{{}^{n+4}C}_4}$
• D. $\frac{3^{n+4}+3^{n+4}}{{{}^{n+4}C}_4}$

Answer 1

Answer: (C)

$\frac{4^{n+4}}{{{}^{n+4}C}_4}$

$\frac{\left({{}^{n}C}_r\right)}{{{}^{r+4}C}_4}=\frac{n!}{r!(n-r)!}\frac{4!r!}{(r+4)!}$
$=\frac{4!}{(n+1)(n+2)(n+3)(n+4)}\left({{}^{n+4}C}_{r+4}\right)=\frac{{{}^{n+4}C}_{r+4}}{{{}^{n+4}C}_4}$
Thus,
$S=\frac{1}{{{}^{n+4}C}_4}[{\sum }_{r=0}^n{3}^{r+4}\left({{}^{n+4}C}_{r+4}\right)+{\sum }_{r=0}^3{{}^{n+4}C}_r{3}^r]$
$=\frac{1}{{{}^{n+4}C}_4}{\sum }_{k=0}^{n+4}{{}^{n+4}C}_k{3}^k=\frac{4^{n+4}}{{{}^{n+4}C}_4}$