Question

Find the value of ${\sum }_{r=0}^{n-1}\frac{{}^n{C}_r}{{}^n{C}_r+{}^n{C}_{r+1}}$ when n=100

• A. 100
• B. 50
• C. 51
• D. 101

Answer 1

The correct option is B

50

We know, ${}^n{C}_r+{}^n{C}_{r-1}={}^{n-1}{C}_r$
$\Rightarrow {}^n{C}_r+{}^n{C}_{r+1}={}^{n+1}{C}_{r+1}$
$\Rightarrow \frac{{}^n{C}_r}{{}^n{C}_r+{}^n{C}_{r+1}}=\frac{{}^n{C}_r}{{}^{n+1}{C}_{r+1}}$
=$\frac{n!}{(n-r)!\times r!}\times \frac{(r+1)!\times (n-r)!}{(n+1)!}$

= $\frac{r+1}{(n+1)}$
$\Rightarrow {\sum }_{r=0}^{n-1}\frac{{}^n{C}_r}{{}^n{C}_r+{}^n{C}_{r+1}}={\sum }_{r=0}^{n-1}\frac{r+1}{(n+1)}$
=$\frac{1}{(n+1)}(\frac{n(n-1)}{2}+n)$
=$\frac{1}{(n+1)}\left(\frac{n(n-1)+2n}{n}\right)$
=$\frac{1}{(n+1)}\left(\frac{n^2+n}{2}\right)$
=$\frac{1}{(n+1)}\left(\frac{n(n+1)}{2}\right)$
=$\frac{n}{2}\Rightarrow$ When n=100, answer = $\frac{100}{2}$ =50