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

In summary, the binomial identity $\sum_{j=0}^{n}(-1)^j{n \choose j}=0$ can be proven in two different ways: through an induction proof and through another approach that has not been mentioned. The induction proof has not been provided, but could potentially be an easy solution.
  • #1
lfdahl
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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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  • #2
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.
 
  • #3
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.
 
  • #4
Thankyou very much, Olinguito, for your participation and two nice approaches! (Yes)
 

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

What is the binomial identity ∑(-1)^j(n choose j)=0?

The binomial identity ∑(-1)^j(n choose j)=0 is a mathematical expression that represents the sum of alternating binomial coefficients, where n is a positive integer. It is also known as the Vandermonde's identity.

What is the significance of the binomial identity?

The binomial identity has many applications in combinatorics, probability, and algebra. It is often used to simplify complicated expressions and to prove other mathematical identities.

How can the binomial identity be proved?

The binomial identity can be proved using mathematical induction or by using the properties of binomial coefficients. It can also be proved by using the binomial theorem or by using combinatorial arguments.

What is the role of the (-1)^j term in the binomial identity?

The (-1)^j term in the binomial identity alternates between positive and negative values as j increases. This helps to cancel out the terms in the sum, resulting in a final value of 0.

Can the binomial identity be extended to other values of n?

Yes, the binomial identity can be extended to any real or complex value of n using the generalized binomial theorem. However, for the identity ∑(-1)^j(n choose j)=0 to hold, n must be a positive integer.

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