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Find:$$\sum_{k=1}^{n} \binom{n}{k}k^{r}$$

For r=0 the sum is obviously $2^{n}$.

For r=1 the sum is $n2^{n-1}$.

For r=2 the sum is $n(n+1)2^{n-2}$.

Here's what I've tried:

$$\frac{d(1+x)^{n}}{dx}=n(1+x)^{n-1}=\binom{n}{1}+2\binom{n}{2}x+3\binom{n}{3}x^{2}+\cdots+n\binom{n}{n}x^{n-1}$$

$$\frac{d[x\cdot n(1+x)^{n-1}]}{dx}=\binom{n}{1}+2^{2}\binom{n}{2}x+3^{2}\binom{n}{3}x^{3}+\cdots+n^{2}\binom{n}{n}x^{n-1}$$

So if we continue like this and put x=1 we will get our result.However I cannot generalize this.

I think the answer has something to do with the Stirling Numbers of the Second kind.

Another way to present this question: Find $A_{1},A_{2},\cdots ,A_{r}$ such that $$k^{r}=A_{1}k+A_{2}k(k-1)+A_{3}k(k-1)(k-2)+\cdots$$

If someone can answer the second question I can solve the first one.

Shaswata
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1 Answers1

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You're right that it has to do with Stirling numbers. Wikipedia's article on Stirling numbers of the second kind answers your second question:

If we let $$(x)_n=x(x-1)(x-2)\cdots(x-n+1) \,$$ (in particular, $(x)_0 = 1$ because it is an empty product) be the falling factorial, we can characterize the Stirling numbers of the second kind by $$ \sum_{k=0}^n \left\{\begin{matrix} n \\ k \end{matrix}\right\}(x)_k=x^n$$

Peter Taylor
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