# Binomial alternating series- even numbers

QuantumP7

## Homework Statement

Ok, I know that $$(-1)^r \binom {n} {r}$$ is supposed to equal 0. But I have plugged some numbers into this series, and this doesn't seem to be true for even numbers of n? Like for n = 4 and r = 4, I have:

$$1 - \frac{4!}{1!3!} + \frac{4!}{2!2!} - \frac{4!}{3!1!} + \frac{4!}{4!}$$. The $$\frac{4!}{2!2!}$$ seems to be in the way. For even numbers, the $$\frac{n!}{(n/2)!(n/2)!}$$ does not seem to cancel out, resulting in the series not being equal to zero. Am I doing something wrong at $$\frac{n!}{(n/2)!(n/2)!}$$?

## Homework Equations

$$1 - \binom {n} {1} + \binom {n} {2} - \binom {n} {3} + \cdots + (-1)^r \binom {n} {n} = 0$$

## The Attempt at a Solution

See #1.

Homework Helper
Hi QuantumP7!

You're expanding (1 - 1)n sing the binomial theorem, so obviously it has to be zero?

Yes it is … your example for n = 4 goes 1 -4 6 -4 1 = 8 - 8 = 0.

Write out a Pascal's triangle and check as many rows as you like!

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$$1 - \frac{4!}{1!3!} + \frac{4!}{2!2!} - \frac{4!}{3!1!} + \frac{4!}{4!}$$. The $$\frac{4!}{2!2!}$$ seems to be in the way. For even numbers, the $$\frac{n!}{(n/2)!(n/2)!}$$ does not seem to cancel out, resulting in the series not being equal to zero. Am I doing something wrong at $$\frac{n!}{(n/2)!(n/2)!}$$?

It seems to be in the way of what? The other terms in that series don't cancel out in any natural way.

When you have an even number of terms, expanding (1-1)n is trivial because you just get an n choose k and a negative n choose k for each k. When n is odd though you get something that's genuinely interesting

Last edited:
QuantumP7
I get it now! Thank you two so much!

Homework Helper

## Homework Statement

Ok, I know that $$(-1)^r \binom {n} {r}$$ is supposed to equal 0. But I have plugged some numbers into this series, and this doesn't seem to be true for even numbers of n? Like for n = 4 and r = 4, I have:

$$1 - \frac{4!}{1!3!} + \frac{4!}{2!2!} - \frac{4!}{3!1!} + \frac{4!}{4!}$$.
For n odd "like" terms such as
$$\begin{pmatrix}5 \\1\end{pmatrix}$$
and
$$\begin{pmatrix} 5 \\ 4\end{pmatrix}$$
will have opposite sign and cancel. You seem to think the same happens with even n. It doesn't:

$\begin{pmatrix}4 \\ 1\end{pmatrix}= \frac{4!}{1!3!}$
and

$\begin{pmatrix}4 \\ 3\end{pmatrix}= \frac{4!}{3!1!}$
have the same sign and add
the "middle term" is necessary to cancel all the other terms

The $$\frac{4!}{2!2!}$$ seems to be in the way. For even numbers, the $$\frac{n!}{(n/2)!(n/2)!}$$ does not seem to cancel out, resulting in the series not being equal to zero. Am I doing something wrong at $$\frac{n!}{(n/2)!(n/2)!}$$?
Did you actually do that calculation? It is 1- 4+ 6- 4+ 1= -3+ 6- 3= 0.

## Homework Equations

$$1 - \binom {n} {1} + \binom {n} {2} - \binom {n} {3} + \cdots + (-1)^r \binom {n} {n} = 0$$

See #1.