Let ‘X’ be a random variable having binomial distribution with parameters $(n>1) \text{and }p(0<p<1)\text{. Then }E\left(\frac{1}{1+X}\right) \text{equals}$

$x~B\cdot D(n,p),0<p<1$

$y=\left(\frac{1}{1+x}\right);x=0,1,2,\dots \dots ,n$

$y=1\because \frac{1}{2},\frac{1}{3},\dots \dots ,\frac{1}{n+1}$

$E\left(y\right)=\sum _{r=0}^n\left[\frac{1}{1+r}{\times }^n{C}_r{q}^{n-r}{p}^r\right]$

$=\frac{1}{(n+1)p}\sum _{r=0}^n{[}^{n+1}{C}_{r+1}{q}^{n-r}{p}^{r+1}]\left(\right(n-r)=(n+1)-(r+1\left)\right)$

$=\frac{1}{(n+1)p}\left[\right(q+p{)}^{n+1}{-}^{n+1}{C}_0{q}^{n+1}];(p+q=1)=\frac{[1-(1-p{)}^{n+1}]}{(n+1)p}$