I've got a new exercise. $$ \sum_{k=0}^{n} \frac{\binom{n}{k}}{n2^n+k} $$ I've tried this: $$ (1+x)^n=\sum_{k=0}^n\binom{n}{k}1^{n-k}x^k $$ Substitute $$ x=x^{n2^n} $$ And integrate from 0 to 1 with respect to x. $$ \int_0^1(1+x^{n2^n})^n = 1+\frac{1}{n2^n+1}\binom{n}{1}+\frac{1}{n2^n+2}\binom{n}{2}+...++\frac{1}{n2^n+n}\binom{n}{n} | -1+\frac{1}{n2^n} $$
$$ \int_0^1(1+x^{n2^n})^n -1+\frac{1}{n2^n} = \sum_{k=0}^{n} \frac{\binom{n}{k}}{n2^n+k} $$ And from here i have no idea!Thank you for your time!