# Binomial theorem, prove an expression

Forgive me guys, i don't really know how to edit this so it would look like 'maths' but i really don't understand what this is asking me to do -_-

Use the Binomial Theorem to show that:

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

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HINT: Look at $(1-1)^n$ ${}{}$

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but when n is 0, we get 0 to the power of 0 which is undefined – Supernatural Feb 28 '13 at 1:31
Since the proposed formula is true only for $n \ge 1$, this "not defined" case is irrelevant. – GEdgar Feb 28 '13 at 1:33
I've worked this out on paper - it does work but i don't understand why we can safely say that 0 = (1-1)^n – Supernatural Feb 28 '13 at 1:38
@DoctorR $(1-1) = 0$ and $0^n = 0$ for all $n \in \mathbb{R} \backslash \{0\}$. – user17762 Feb 28 '13 at 1:45
but k starts from 0, doesn't that mean that n = 0? – Supernatural Feb 28 '13 at 1:46

Hint: How would you use binomial theorem to expand $(1+ (-1))^n$

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I see, but why is it safe to say that 0 = (1+(-1))^n? values of k start at 0 so n could be 0 couldn't it? – Supernatural Feb 28 '13 at 1:41
The statement doesnt really make sense for n=0 but for all n>0 it works – Sean Ballentine Feb 28 '13 at 1:43
ok thank you!!! – Supernatural Feb 28 '13 at 1:44