Let X∼Binomial(n,p)X \sim \text{Binomial}(n, p)X∼Binomial(n,p). Find the expected value of 1X+1\frac{1}{X+1}X+11.
1−(1−p)n+1(n+1)p\frac{1 - (1-p)^{n+1}}{(n+1)p}(n+1)p1−(1−p)n+1
1(n+1)p\frac{1}{(n+1)p}(n+1)p1
1−(1−p)nnp\frac{1 - (1-p)^n}{np}np1−(1−p)n
\frac{(1-p)^{n+1}}{(n+1)p}