sum equals to...?

Alen0905

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Mar 22, 2020
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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

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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

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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

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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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