Question

The value of $\sum _{r=0}^{n-1}\frac{{}^n{C}_r}{{}^n{C}_r{+}^n{C}_{r+1}}$ is equals to ..............

• A. n + 1
• B.
• C. n + 2
• D. n

Answer 1

The correct option is B

Let's try to change this expression into a general expression so that we can simplify it.
We know the general expression $\frac{{}^n{C}_{r+1}}{{}^n{C}_r}=\frac{n-r}{r+1}$

So, now we will divide numerator and denominator by ${}^n{C}_r$ to get a generalized expression.
$\sum _{r=0}^{n-1}\frac{{}^n{C}_r}{{}^n{C}_r+{}^n{C}_{r+1}}=\sum _{r=0}^{n-1}\frac{1}{1+\frac{{}^n{C}_{r+1}}{{}_r^C}}=\sum _{r=0}^{n-1}\frac{1}{1+\frac{n-r}{r+1}}=\sum _{r=0}^{n-1}\frac{1}{\frac{r+1+n-r}{r+1}}$

$=\sum _{r=0}^{n-1}\frac{r+1}{n+1}=\frac{1}{n+1}\sum _{r=0}^{n-1}(r+1)$
$=\frac{1}{(n+1)}[1+2+.....+n]$
But we know, Sum of the first of the first n terms = $\frac{n(n+1)}{2}$
Therefore,
$=\frac{n(n+1)}{(n+1)2}=\frac{n}{2}$