Binomial theorem, prove an expression

Question

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:

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

Your formula is only for $n>0$, because if $n=0$ then the right-hand side is $1$. — Mark, Mar 02 '17 at 21:00

score 1 · Accepted Answer · answered Feb 28 '13 at 01:28

1

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

answered Feb 28 '13 at 01:28

but when n is 0, we get 0 to the power of 0 which is undefined – Supernatural Feb 28 '13 at 01:31
2

Since the proposed formula is true only for $n \ge 1$, this "not defined" case is irrelevant. – GEdgar Feb 28 '13 at 01:33
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 01:38
@DoctorR $(1-1) = 0$ and $0^n = 0$ for all $n \in \mathbb{R} \backslash {0}$. – Feb 28 '13 at 01:45
but k starts from 0, doesn't that mean that n = 0? – Supernatural Feb 28 '13 at 01:46
@DoctorR, where the sum starts is irrelevant, you are interested in where it ends. And if the case $n = 0$ is of interest, handle it separately. – vonbrand Feb 28 '13 at 19:09

score 0 · Answer 2 · answered Feb 28 '13 at 01:28

0

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

answered Feb 28 '13 at 01:28

Sean Ballentine

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 01:41
The statement doesnt really make sense for n=0 but for all n>0 it works – Sean Ballentine Feb 28 '13 at 01:43

2 Answers2