Prove Binom Sum $\sum_{k=0}^n(-1)^k \binom{n}{k} = 0$

Question

Let: $$ (-1)^0=1 $$ I need to prove that: $$ \sum_{k=0}^n(-1)^k \binom{n}{k} = 0 $$

Thanks!

Answer 1

Hint: Expand $(1-1)^n$ using binomial theorem.

Answer 2

Hint: $(-1)^k =(-1)^k\cdot 1^{n-k}$

Answer 3

According to Binomial theorem we have : $$\sum_{k=0}^{n}\binom{n}{k} a^{n-k}b^k=(a+b)^n$$

Thus we can make a special case $$0=(1-1)=(1-1)^n=\sum_{k=0}^{n}\binom{n}{k} 1^{n-k}(-1)^k=\sum_{k=0}^{n}\binom{n}{k} (-1)^k$$ which is what you were looking for.

Answer 4

In Pascal's triangle, every element it the sum of the two elements above it. So if you sum a row with alternating signs, every element of the row above appears two times with opposite signs and everything telescopes.

Example:

$$1\ \ \ \ \ \ \ \ \ \ 3\ \ \ \ \ \ \ \ \ \ 3\ \ \ \ \ \ \ \ \ \ 1$$ $$+\ -\ -\ +\ +\ -\ -\ +$$ $$1\ \ \ \ -4\ \ \ \ \ \ \ \ \ 6\ \ \ \ \ -4\ \ \ \ \ \ \ \ \ 1$$