0
$\begingroup$

I recently came across the following sum on Wikipedia: $$\sum_{i=0}^n (-1)^i \binom{n}{i} = 0. \qquad (1)$$ I was wondering what proof exists for $$\sum_{i=1}^n (-1)^i \binom{n}{i} = -1, \quad (2)$$ without assuming $(1)$. So far, I've split up the cases $n$ even/$n$ odd and have been able to prove that, when $n$ is odd, all terms except the last (which simplifies to $-1$) cancel. With $n$ even the cancellations are not as easy or obvious, though I've checked for a few even values of $n$ that $(2)$ still turns out to be $-1$. I'd appreciate a proof that shows that verifies $(2)$.

$\endgroup$
1
$\begingroup$

More generally $\displaystyle\;\sum_{i=0}^n a_i = a_0 + \sum_{\color{red}{i=1}}^n a_i\,$.

In this case $\displaystyle\;a_i = (-1)^i \binom{n}{i}\,$, therefore $\,a_0=1\,$, then $\displaystyle\;\sum_{i=0}^n (-1)^i \binom{n}{i} = 1 + \sum_{\color{red}{i=1}}^n (-1)^i \binom{n}{i}\,$.

It follows by the binomial theorem that $\displaystyle\;\sum_{\color{red}{i=1}}^n (-1)^i \binom{n}{i}=-1 + \sum_{i=0}^n (-1)^i \binom{n}{i} = -1 + (1-1)^n \,$.

$\endgroup$
  • 1
    $\begingroup$ Nice answer, and thanks for making it clear where you got the $(1-1)^n$ term! $\endgroup$ – Linus Rastegar Sep 1 '17 at 8:24
1
$\begingroup$

The first it's Its $$(1-1)^n=0$$ The second it's $$(1-1)^n-1=-1$$

$\endgroup$
1
$\begingroup$

$$(1-1)^n=\binom{n}{0}1^n(-1)^0+\binom{n}{1}1^{n-1}(-1)^{1}++\binom{n}{2}1^{n-2}(-1)^{2}+...++\binom{n}{n}1^{n-n}(-1)^{n}=0^n=0\\\to \\\sum_{i=0}^n (-1)^i \binom{n}{i} = 0. $$now expand sigma for $i=0$ and you will have $$\sum_{i=0}^n (-1)^i \binom{n}{i} =\binom{n}{0}1^n(-1)^0+\sum_{i=1}^n (-1)^i \binom{n}{i} = 0\\1+\sum_{i=1}^n (-1)^i \binom{n}{i} = 0 \to \\\sum_{i=1}^n (-1)^i \binom{n}{i} =0-1$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.