Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I want to compute $$S(n,m,a)=\sum_{k=0}^{n}k^{m}\cdot\binom{n}{k}\cdot a^k.$$ With $n,m\in\mathbb N$, $a\neq0$ and $S(n,0,a)=(a+1)^n$.

What I have found already:

I don't see any other options then integrating until we've got Newton's formula and then differentiating it as many times as we integrated.

I have found some values:$$S(n,1,a)=na\cdot(a+1)^{n-1}$$ $$S(n,2,a)=an(an+1)(a+1)^{n-2}$$ $$S(n,3,a)=an(a^2n^2+3an-a+1)(a+1)^{n-3}.$$

And, since $\displaystyle\sum_{k=0}^{n}k^{m}\cdot\binom{n}{k}a^k=a\cdot\frac{\mathrm d}{\mathrm da}\int\displaystyle\sum_{k=0}^{n}k^{m}\cdot\binom{n}{k}a^{k-1}\mathrm da=a\cdot\frac{\mathrm d}{\mathrm da}\sum_{k=0}^{n}k^{m-1}\cdot\binom{n}{k}a^k$, I have the recursion $$S(n,m,a)=a\cdot\frac{\mathrm d}{\mathrm da}S(n,m-1,a).$$ But I can't find any general formula.

share|improve this question
The verb corresponding to the noun "derivative" is "differentiate". –  joriki Oct 29 '12 at 15:32
If your reasonning is correct, then $S(n,2,a)=a\cdot\frac{\mathrm d}{\mathrm da}S(n,1,a)=na(n-1)(a+1)^{n-2}$ –  wnvl Oct 29 '12 at 17:19
Just a small remark, I think you forgot the summation just before ",I have the recursion" –  wnvl Oct 29 '12 at 17:22
Maybe it might help that $$S(n,m,a_1)=n\cdot\frac{\mathrm d}{\mathrm da_1}\left(\left.\frac{\mathrm d}{\mathrm da_2}\left(...\left.\frac{\mathrm d}{\mathrm da_m}\left(\left(1+\prod_{k=1}^{m}a_k\right)^{n-1}\cdot\prod_{k=1}^{m}a_k\right)‌​\right|_{a_m=1}...\right)\right|_{a_2=1}\right)$$ –  barto Oct 29 '12 at 18:24
@wnvl: The reasoning is right, but $S(n,1,a)=na(a+1)^{n-1}$. –  barto Oct 29 '12 at 19:12
add comment

1 Answer

up vote 1 down vote accepted

The following is not a complete solution. The sum is reduced to a sum over $m$ rather than $n+1$ terms. Let $D = \partial/\partial a$. Then $$\begin{eqnarray*} S(n,m,a) &=& \sum_{k=0}^n {n\choose k} k^m a^k \\ &=& (a D)^m \sum_{k=0}^n {n\choose k} a^k \\ &=& (a D)^m (1+a)^n \\ &=& \left(\sum_{k=1}^m \left\{m\atop k\right\} a^k D^k\right)(1+a)^n \\ &=& \sum_{k=1}^m \left\{m\atop k\right\} a^k \frac{n!}{(n-k)!} (1+a)^{n-k} \end{eqnarray*}$$ Therefore, $$S(n,m,a) = n!(1+a)^n \sum_{k=1}^m \left\{m\atop k\right\} \frac{1}{(n-k)!} \left(\frac{a}{1+a}\right)^k.$$ The operator $(a D)^m$ and its relation to the Stirling numbers of the second kind is dealt with here.

share|improve this answer
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.