basic sum transformations

Question

I am new to math and I've a lot of problems in basic transformations of equations. For example this one:

$$ {n^n}-\sum_{k=1}^n {n \choose k} {(-1)^{k-1}n^{n-k}} = \sum_{k=0}^n {n \choose k} {(-1)^{k}n^{n-k}} = (n-1)^n$$

I don't know how $n^n$ will be integrated !?

Note that, $n^n$ is integrated, because $k$ starts from $0$ on the right hand side of the equation instead of $1$ and also notice that ${n \choose 0}=1$. — Mhenni Benghorbal, Apr 14 '13 at 18:07
Yeah, expanding on Mhenni's answer, this looks to be the binomial theorem split up a little. $n$ is constant. — AlexM, Apr 14 '13 at 18:08
Well... I do not see the problem. The term corresponding to $k=0$ in the sum $\sum_{k=0}^n {n \choose k} {(-1)^{k}n^{n-k}}$ is just ${n \choose 0}{(-1)^{0}n^{n}}=n^n$... — Simone, Apr 14 '13 at 18:10

score 0 · Answer 1 · answered Apr 14 '13 at 18:51

0

Okay, understand: $(-1)^{k-1}$ is only to change the parity of the next summand. It is also possible to write:

$$ {n^n}+\sum_{k=1}^n {n \choose k} {(-1)^{k}n^{n-k}} = \sum_{k=0}^n {n \choose k} {(-1)^{k}n^{n-k}} = (n-1)^n$$

answered Apr 14 '13 at 18:51

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