Help in binomial theorem

**John G** · January 11th 2009, 06:58 PM

How can i calculate this sum?

Thanks for any kind of help.

**ThePerfectHacker** · January 11th 2009, 09:30 PM

Remember, $(1+x)^m = \sum_{j=0}^m x^j {m\choose j}$ .

This sum can be written as, $\sum_{k=0}^n \sum_{i=0}^k \frac{n!}{k!(n-k)!}\cdot \frac{k!}{i! (k-i)!} (-3)^i = \sum_{k=0}^n \sum_{i=0}^k {n\choose k}\boxed{ {k\choose i} (-3)^i }$

Using the formula above on boxed term we get, $\sum_{k=0}^n{n\choose k} (-2)^k = (-1)^n$

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