Prove series result?

**fardeen_gen** · May 21st 2009, 10:21 AM

If $S_p = 1^p + 2^p + 2^p + \mbox{...} + n^p$ , prove that $S_0 + (p + 1)S_1 + \frac{p^2 + p}{2}S_2 + \mbox{...} + (p + 1)S_p = (n + 1)^{p + 1} - 1$

**TheAbstractionist** · May 21st 2009, 01:05 PM

Originally Posted by fardeen_gen

If $S_p = 1^p + 2^p + \color{red}3\color{black}^p + \mbox{...} + n^p$ , prove that $S_0 + (p + 1)S_1 + \frac{p^2 + p}{2}S_2 + \mbox{...} + (p + 1)S_p = (n + 1)^{p + 1} - 1$

Hi fardeen_gen.

By the binomial theorem,

$(k+1)^{p+1}-k^{p+1}\ =\ 1+(p+1)k+\frac{(p+1)p}2k^2+\cdots+(p+1)k^p$

Now sum both sides from $k=1$ to $n$ and you have your answer.

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