Alternating sum of binomial coefficients equal to $1$

Question

I have to prove that $\sum_{k=1}^{n} (-1)^{k+1}\binom{n}{k}=1.$ I tried to prove that by induction, with the base case trivial.

By the binomial coefficient property we have that $\binom{n}{k}=\binom{n-1}{k}+\binom{n-1}{k}$.

So if we add the summatory and separate the n term in the summatory we have that:

$\sum_{k=1}^{n} (-1)^{k+1}\binom{n}{k}=(-1)^{n+1}\binom{n}{n}+\sum_{k=1}^{n}(-1)^{k+1}\binom{n-1}{k-1}+\sum_{k=1}^{n}(-1)^{k+1}\binom{n-1}{k-1}=(-1)^{n+1}+1+0\neq1$.

(Where the last equality comes from induction hypotesis and by the property $\sum_{k=0}^{n}(-1)^{k}\binom{n}{k}=0$)

Where am i wrong?

First of all $\binom{n}{k}\color{red}{\neq} \binom{n-1}{k}+\binom{n-1}{k}$. For $n=3$ and $k=2$ we have $\binom{3}{2}\neq \binom{2}{2}+\binom{2}{2}\Rightarrow 3\neq 1+1$ The right identity is $\binom{n}{k} = \binom{n-1}{k} + \binom{n-1}{k-1}$ — callculus42, Aug 20 '20 at 20:56
If you know the property $\sum_{k=0}^{n}(-1)^{k}\binom{n}{k}=0$ (as you say at the end), the relationship you want to establish is immediate because $\binom{n}{0}=1$! — Jean Marie, Aug 20 '20 at 21:02
@Andrea Licata I think that you showed good instinct in making your first try induction rather than elegant algebraic manipulation. This particular problem is unusual however; as shown by user's answer, the binomial theorem (https://en.wikipedia.org/wiki/Binomial_theorem) leads to a direct demonstration. — user2661923, Aug 20 '20 at 21:07
Does this answer your question? Alternating sum of binomial coefficients: given $n \in \mathbb N$, prove $\sum^n_{k=0}(-1)^k {n \choose k} = 0$ — Martin R, May 11 '23 at 19:18
Also: https://math.stackexchange.com/q/94514/42969 (of which this question had been closed as a duplicate). — Martin R, May 12 '23 at 08:55

score 2 · Answer 1 · 2020-08-20T21:24:25.340

Hint: If you know that if $n$ is nonnegative integer, then $$(x+y)^{n}=\sum_{k}\binom{n}k x^{k}y^{n-k}$$Now, if $x=-1$ and $y=1$, we find that the alternating sum across any row of Pascal's triangle is zero, except of course for the top row: $$\boxed{\sum_{k}(-1)^{k} \binom{n}k =\left\{ \begin{align*} 0, \quad n\geq 1\\ 1, \quad n=0 \end{align*} \right. }.$$

This is obvious from the symmetry relation when $n$ is odd, but less clear when $n$ is even.

Hint (induction): We prove this identity by induction on $n$. For $n=0$, it's clear.

score 0 · Answer 2 · answered Aug 20 '20 at 21:01

0

By definition

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

therefore

$$\sum_{k=1}^{n}(-1)^{k}\binom{n}{k}=-1 \implies \sum_{k=1}^{n}(-1)^{k+1}\binom{n}{k}=1$$

answered Aug 20 '20 at 21:01

user

154,566

How does this differ from the solutions given in https://math.stackexchange.com/q/94514/42969? It is unclear to me why you reopened the question. – Martin R May 12 '23 at 08:56
@MartinR It is three years ago question but it seems that the reason is that here the OP is seeking for a specific issue. OP is aware of the result $\sum_{k=0}^{n}(-1)^{k}\binom{n}{k}=0$ and is asking why $\sum_{k=1}^{n}(-1)^{k+1}\binom{n}{k}=1$. – user May 12 '23 at 09:28

Alternating sum of binomial coefficients equal to $1$

$\sum_{k=1}^{n} (-1)^{k+1}\binom{n}{k}=(-1)^{n+1}\binom{n}{n}+\sum_{k=1}^{n}(-1)^{k+1}\binom{n-1}{k-1}+\sum_{k=1}^{n}(-1)^{k+1}\binom{n-1}{k-1}=(-1)^{n+1}+1+0\neq1$.

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