Prove $\sum_{0}^{n}(-1)^i\binom{n}{i} = 0$

Question

I have to prove that $\sum_{0}^{n}(-1)^i\binom{n}{i} = 0$,

I know i have to use the binomial theorem,But i dont know how.

Thanks.

Possible duplicate of Alternating sum of binomial coefficients — , Mar 15 '16 at 20:19
Possible duplicate of Proving $\sum\limits_{k=0}^{n} (-1)^{k} \binom{n}{k} = 0$ — colormegone, Mar 15 '16 at 23:28

score 3 · Answer 1 · answered Mar 15 '16 at 18:05

3

$$ 0 = \bigl((1+ (-1)\bigr)^n = \sum_{i=0}^n \binom ni 1^{n-i}(-1)^i = \sum_{i=0}^n \binom ni (-1)^i $$

answered Mar 15 '16 at 18:05

martini

score 3 · Answer 2 · answered Mar 15 '16 at 18:06

The Binomial Theorem: $$(x+y)^n = \sum\limits_{i=0}^n \binom{n}{i}x^{n-i}y^i$$

For example $(x+y)^4 = x^4+4x^3y+6x^2y^2+4xy^3+y^4$

Using the above, notice what happens when $x=1$ and $y=-1$

$0=(1-1)=(1+(-1))^n=\sum\limits_{i=0}^n\binom{n}{i}1^{n-i}(-1)^i=\sum\limits_{i=0}^n\binom{n}{i}(-1)^i$

In general, with other sums like this, one can throw in additional factors of $1$ to an arbitrary power to notice the pattern.

For example, $\sum\limits_{i=0}^n\binom{n}{i} = \sum\limits_{i=0}^n\binom{n}{i}(1)^i(1)^{n-i} = (1+1)^n = 2^n$

2 Answers2