Question

${\sum }_{r=0}^{n-1}\frac{{{}^{n}C}_r}{{{}^{n}C}_r+{{}^{n}C}_{r+1}}$ is equal to

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

Answer 1

The correct option is A $\frac{n}{2}$

${\sum }_{r=0}^{n-1}$$\frac{{}^n{C}_r}{{}^n{C}_r{+}^n{C}_{r+1}}$

=${\sum }_{r=0}^{n-1}$$\frac{{}^n{C}_r}{{}^{n+1}{C}_{r+1}}$ ----------(because ${}^n{C}_r{+}^n{C}_{r-1}{=}^{n+1}{C}_r$)

=${\sum }_{r=0}^{n-1}\frac{{}^n{C}_r}{{\frac{(n+1)}{(r+1)}}^n{C}_r}$

=${\sum }_{r=0}^{n-1}\frac{r+1}{n+1}$

=$\frac{n(n+1)}{2(n+1)}$

=$\frac{n}{2}$

hence option ${}^{\prime }{A}^{\prime }$ is correct.