MHB Prove the binomial identity ∑(-1)^j(n choose j)=0

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The discussion centers on proving the binomial identity ∑(-1)^j(n choose j)=0 using two different methods. One participant suggests using mathematical induction as a potential proof strategy. Another participant expresses surprise at the challenge level, indicating that they initially thought it was an easier problem. They also appreciate the contributions of another member who provided two approaches to the proof. The conversation highlights the collaborative nature of problem-solving in mathematics.
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Prove the binomial identity:

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

- in two different ways
 
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lfdahl said:
Prove the binomial identity:

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

- in two different ways
The only way I can think of is to do an induction proof. I haven't sat down to do it but it shouldn't be too hard.

-Dan

(Ahem!) I thought you were asking for help. When I saw it was a challenge I couldn't edit it out. (I thought it was a rather easy problem for you to be asking for help on.) Anyway, if someone else doesn't post it I'll get back to it later.
 
First method.

When $n$ is odd, it’s easy. The coefficients $\displaystyle\binom nj$ and $\displaystyle\binom n{n-j}$ from $j=0$ to $j=n$ pair up nicely; also $(-1)^{n-j}=(-1)^n(-1)^j=-(-1)^j$ (as $n$ is odd). Thus
$$\sum_{j=0}^n(-1)^j\binom nj$$

$\displaystyle=\ \sum_{j=0}^{\frac{n-1}2}\left[(-1)^j\binom nj + (-1)^{n-j}\binom n{n-j}\right]$

$\displaystyle=\ \sum_{j=0}^{\frac{n-1}2}\left[(-1)^j\binom nj - (-1)^j\binom n{n-j}\right]$

$=\ 0$.

Suppose $n$ is even, so $(-1)^n=1$ and $(-1)^{n-1}=-1$. We use the well-known identity
$$\binom nj\ =\ \binom{n-1}{j-1}+\binom{n-1}j$$
(which is simply saying that a binomial coefficient is the sum of the two coefficients above it in Pascal’s triangle). Then
$$\sum_{j=0}^n(-1)^j\binom nj$$

$\displaystyle=\ 1\,+\,\sum_{j=1}^{n-1}\left[(-1)^j\binom{n-1}{j-1} + (-1)^j\binom {n-1}j\right]\,+\,(-1)^n$

$\displaystyle=\ 1\,+\,\sum_{j=1}^{n-1}\left[-(-1)^{j-1}\binom{n-1}{j-1} + (-1)^j\binom {n-1}j\right]\,+\,1$

$\displaystyle=\ 1\,+\,\left[-(-1)^0\binom{n-1}0+(-1)^{n-1}\binom{n-1}{n-1}\right]\,+\,1$ (by telescoping)

$=\ 1+[(-1)+(-1)]+1$

$=\ 0$.Second method.

Expand $0=[1+(-1)]^n$ binomially.
 
Thankyou very much, Olinguito, for your participation and two nice approaches! (Yes)
 
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