sum equals to...?

[Alen0905]

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

 

[MarkFL]

Hello and welcome to FMH!

Consider the following application of the binomial theorem:

[MATH]3^n=(2+1)^n=\sum_{k=0}^n\left({n \choose k}2^{k}1^{n-k}\right)[/MATH]

 

[Alen0905]

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.

 

[Steven G]

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 [MATH]2^n=(1+1)^n=\sum_{k=0}^n\left({n \choose k}1^{k}1^{n-k}\right)[/MATH] =[MATH]\sum_{k=0}^n\left({n \choose k}\right)[/MATH]