# sum equals to...?

#### Alen0905

##### New member
apparently the sum of the binomials equals 2^n...but multiplied with what? its confusing me... ...would appreciate help. Thanks

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#### MarkFL

##### Super Moderator
Staff member
Hello and welcome to FMH!

Consider the following application of the binomial theorem:

$$\displaystyle 3^n=(2+1)^n=\sum_{k=0}^n\left({n \choose k}2^{k}1^{n-k}\right)$$

#### Alen0905

##### New member
that's awesome...i figure my biggest problem in math is perspective, often i delude myself and make problems harder in my mind then they are in reality.

#### Jomo

##### Elite Member
apparently the sum of the binomials equals 2^n...but multiplied with what? its confusing me... ...would appreciate help. Thanks
It is great that you like Mark's post.

Taken your question literally the 2^n is not multiplied by anything.

Consider $$\displaystyle 2^n = (1+1)^n$$ .........[edited ... to the corner] Now using the binomial expansion to $$\displaystyle (1+1)^n$$ we get $$\displaystyle 2^n=(1+1)^n=\sum_{k=0}^n\left({n \choose k}1^{k}1^{n-k}\right)$$ =$$\displaystyle \sum_{k=0}^n\left({n \choose k}\right)$$

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