I would like to calculate, or bound from above, the following sum $$ \sum_{i=0}^n(n2i)^p{p \choose i}, $$ here $p\geq 2$.
Any references are very welcome.
Thank you.
I would like to calculate, or bound from above, the following sum $$ \sum_{i=0}^n(n2i)^p{p \choose i}, $$ here $p\geq 2$. Any references are very welcome. Thank you. 


On of methods for finding an approximation or bound for a finite sum is using the following formula $$ \sum_{n=a}^b f(n) \sim \int_a^b f(x)\,dx + \frac{f(a) + f(b)}{2} + \sum_{k=1}^\infty \,\frac{B_{2k}}{(2k)!}\left(f^{(2k  1)'}(b)  f^{(2k  1)'}(a)\right)$$ and by taking $f(i)= (n2i)^p\binom {p}{i}$ you can consider decimal digits of right hand side which are faster than of left hand side. Also $B_k$ here are Bernoulli numbers. Moreover, the sharp bounds of Bernoulli numbers has been computed ,(see here ) So you can also try to find a bound for your sum. 

