show that ${n \choose 0} - {n \choose 1} + \cdots + (-1)^n {n \choose n} = 0$

Question

I was asked to show that this formula is true.

${n \choose 0} - {n \choose 1} + \cdots + (-1)^n {n \choose n} = 0$

I've already proven that ${n \choose 0} + {n \choose 1} + \cdots + {n \choose n} = 2^n$

However, I'm not really sure if these two are related at all or if I'm missing the point here. How do I show that this formula is true?

If you used induction, did you use Pascal's identity? Because Pascal's identity can be used to prove both. Don't forget for this identity, you have to use $n=1$ as the base case, not $n=0$ because it fails. — zhuli, Sep 29 '20 at 13:09

Surb · Answer 1 · 2020-09-29T13:07:42.073

3

Hint

$$\binom{n}{k}=\binom{n}{n-k},$$ and see what happen when $n$ is odd and $n$ is even.

edited Sep 29 '20 at 13:07

answered Sep 29 '20 at 13:00

Surb

When k is odd and n even? – 23408924 Sep 29 '20 at 13:09
no, when $n$ is odd, the result is obvious, but when $n$ is even, it needs more work. – Surb Sep 29 '20 at 13:12
oh I see, thanks! – 23408924 Sep 29 '20 at 13:17

score 3 · Accepted Answer · answered Sep 29 '20 at 13:03

3

The question is the binomial expansion of $(1-1)^n$ which is equal to zero.

answered Sep 29 '20 at 13:03

Anwesha1729

2 Answers2