# Series proof.

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• Sep 13th 2009, 06:15 PM
akolman
Sum proof.
Hello, I am stuck with the following problem.

Verify the following identity for $\displaystyle n \geq 2$.

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

I tried an induction proof, but I couldn't get it right. Do I need to consider when n is odd and when n is even?

Thanks in advance
• Sep 13th 2009, 07:29 PM
Krizalid
this is a sum, not a series, and the solution is quite easy:

$\displaystyle \sum\limits_{k=0}^{n}{\binom nk(-1)^{k}1^{n-k}}=(1-1)^{n}=0^{n}=0.$
• Sep 13th 2009, 09:51 PM
Prove It
Quote:

Originally Posted by akolman
Hello, I am stuck with the following problem.

Verify the following identity for $\displaystyle n \geq 2$.

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

I tried an induction proof, but I couldn't get it right. Do I need to consider when n is odd and when n is even?

Thanks in advance

Like Krizalid said, this is an example of the Binomial Theorem.

If you know that

$\displaystyle (x + y)^n = \sum_{k = 0}^n {\left(_k^n\right) x^{n - k} y^k}$

it is relatively easy to see that if you let $\displaystyle x = 1$ and $\displaystyle y = -1$ you get

$\displaystyle (1 - 1)^n = \sum_{k = 0}^n {\left(_r^n\right) (1)^{n - k} (-1)^k}$

$\displaystyle 0^n = \sum_{k = 0}^n {\left(_k^n\right) (-1)^k}$

Therefore $\displaystyle \sum_{k = 0}^n {\left(_k^n\right) (-1)^k} = 0$.