By setting x equal to the appropriate values in the binomial expansion

Question

I have this equation,

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

the answer is 0 derived by,

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

How come $$ \sum_{k=0}^n (-1)^k {n \choose k} $$ is same as $$ (1-1)^n $$ ??

Thanks,

Answer 1

The binomial theorem states that $$(x+y)^n=\sum_{k=0}^n{n\choose k}x^ky^{n-k}$$.

Now substitute $x=1$ and $y=-1$.

Answer 2

For $n$ odd, it should be obvious: you are adding and subtracting successive binomial coefficients in Pascal's triangle. If $n=3$ your sum is $1-3+3-1$, where each binomial coefficient occurs twice, once with a $+$ sign and once with a $-$ sign.

For even $n$, consider that each entry in row 4 (say) is the sum of two entries in row 3, by the well known recursion. Your sum for $n=4$ is $1-4+6-4+1$ which we should think of as $(0+1) - (1+3) + (3+3) - (3+1) - (1+0)$ which shows the binomial coefficients from the $n=3$ row canceling.