# Prove: \sum_{r = 0}^n \binom{n}{r} = 0 and \sum_{r = 0}^n \binom{n}{r} = 2^n Consider (1 -...

## Question:

Prove:

{eq}\sum_{r = 0}^n \binom{n}{r} = 0 {/eq} and {eq}\sum_{r = 0}^n \binom{n}{r} = 2^n {/eq}

Consider {eq}(1 - 1)^n {/eq} and {eq}(1 + 1)^n {/eq} or use Pascal's equation and proof by induction.

## Binomial expansion

When dealing with binomial expansion, we have to general types of expanding formulas that we utilize in order to expand the more complicated function. If we have a function of the form

{eq}(x+y)^n = \binom{n}{0} x^n + \binom{n}{1}x^{n-1}y^1 + ... + \binom{n}{n} y^n {/eq}

Also, we can expand a function of the form

{eq}(x-y)^n = \binom{n}{0} x^n - \binom{n}{1} x^{n-1} (y)^1 + \binom{n}{2} x^{n-2} (y)^2 - ... {/eq}

So, the signs alternate in between each term.

Suppose we let X=1 and Y=1, then we get

{eq}(1+1)^n = \binom{n}{0} 1^n + \binom{n}{1} 1^{n-1} 1^1 + \binom{n}{2} 1^{n-2} 1^2 +...+ \binom {n}{n} 1^n {/eq}

{eq}= \binom{n}{0} + \binom{n}{1} + \binom {n}{2} + ... + \binom{n}{n} {/eq}

{eq}= \sum_{x=0}^{n} {\binom{n}{x} } {/eq}

{eq}=2^n {/eq}

Now let X=1 and Y=1 but utilize the second form we showed above:

{eq}(1-1)^n = \binom{n}{0} 1^n - \binom{n}{1} 1^{n-1} (1)^1 + \binom{n}{2} 1^{n-2} (1)^2 - ...+ \binom {n}{n} 1^n {/eq}

{eq}= \binom{n}{0} - \binom{n}{1} + \binom {n}{2} - ... + \binom{n}{n} {/eq}

{eq}= \sum_{x=0}^{n} {(-1)^x \binom{n}{x} } {/eq}

This is an alternating sum of binomial coefficients which sums up to 0. Thus we get

{eq}\sum_{x=0}^{n} {(-1)^x \binom{n}{x} } = 0 {/eq}