Show that $$ \sum_{m=0}^n (-1)^m \binom{n}{m} = 0$$
I have first difficulty understanding the summation notation: For example what $ \sum_{m=0}^3 (-1)^m \binom{n}{m}$ would mean? I suppose that it means: $$ \sum_{m=0}^2 (-1)^m \binom{n}{m} = (-1)^0 \binom{2}{0} + (-1)^1 \binom{2}{1} + (-1)^2 \binom{2}{2} $$
$$ \sum_{m=0}^2 (-1)^m \binom{n}{m} = + 1 - 2 + 1 =0 $$
If it is such case, it shows that the value of $(-1)^m$ alternate positive when $m$ is even and negative when $m$ is odd.
Therefore the sum when $m$ is even $+$ the sum when $m$ is odd $= 0$ and we can then factor out the constant $+1$ from the first sum and $-1$ from the second sum and then QED.
Is this approach correct? How to write this using sum notation when $m$ even and when $m$ is odd?
Is there a better approach?
Much appreciated.
\binom{n}{k}
. $\endgroup$ – Em. May 27 '17 at 6:42