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



{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.

Answer and Explanation:

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}

Learn more about this topic:

How to Use the Binomial Theorem to Expand a Binomial

from Algebra II Textbook

Chapter 21 / Lesson 16

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