Binomial Help (1 Viewer)

Kingom · Oct 22, 2016

please help. thanks

InteGrand · Oct 22, 2016

Kingom said:
please help. thanks
View attachment 33529

$\noindent We have$

$\begin{align*}\left(2 + \frac{x}{2}\right)^{n} &= \sum_{k = 0}^{n}\binom{n}{k}\frac{x^{k}}{2^{k}}2^{n-k}\\ \Rightarrow \left(\frac{4+x}{2}\right)^{n} &= \sum_{k = 0}^{n}\binom{n}{k}x^{k}2^{n-2k}.\end{align*}$

$\noindent Now divide both sides by $2^{n}$. The $2^{n-2k}$ in the summand becomes $2^{-2k}$, and the LHS becomes $\left(\frac{4+x}{2^{2}}\right)^{n}$. Then put $x=1$, to obtain $\left(\frac{5}{2^{2}}\right)^{n} = \sum_{k = 0}^{n}\binom{n}{k}2^{-2k}$. This implies the result, by moving the $k=0$ term (which is $1$) from the sum to the other side.$

Binomial Help (1 Viewer)

Kingom

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InteGrand

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