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For all n\geq1 :

$$\left(\begin{matrix}2n\\ 0 \end{matrix}\right) +\left(\begin{matrix}2n\\ 2 \end{matrix}\right) +\left(\begin{matrix}2n\\ 4 \end{matrix}\right) + \ldots +\left(\begin{matrix}2n\\ 2k \end{matrix}\right) + \ldots +\left(\begin{matrix}2n\\ 2n \end{matrix}\right) $$

is equal to

$$\left(\begin{matrix}2n\\ 1 \end{matrix}\right) +\left(\begin{matrix}2n\\ 3 \end{matrix}\right) +\left(\begin{matrix}2n\\ 5 \end{matrix}\right) + \ldots +\left(\begin{matrix}2n\\ 2k+1 \end{matrix}\right) + \ldots +\left(\begin{matrix}2n\\ 2n-1 \end{matrix}\right) .$$

MY SOLUTION

Binomial Formula : (a+b)^{2n} = \sum_{k=0}^{2n} \left(\begin{matrix}2n\\ k \end{matrix}\right) a^{k} b^{n-k}

let a=1 , let b=-1 , let n=2

(1+(-1))^{2(n)} = \left(\begin{matrix}2n\\ 0 \end{matrix}\right) -\left(\begin{matrix}2n\\ 1 \end{matrix}\right) +...\mbox{-}\left(\begin{matrix}2n\\ 2n-1 \end{matrix}\right) + \left(\begin{matrix}2n\\ 2n \end{matrix}\right)

(1+(-1))^{2(4)} = \left(\begin{matrix}2(2)\\ 0 \end{matrix}\right)-\left(\begin{matrix}2(2)\\ 1 \end{matrix}\right)+\left(\begin{matrix}2(2)\\ 02 \end{matrix}\right)-\left(\begin{matrix}2(2)\\ 2(2)-1 \end{matrix}\right)+ \left(\begin{matrix}2(2)\\ 2(2) \end{matrix}\right)

0=1-4+6-4+1

0=0

Thus for all n\geq1 :

\left(\begin{matrix}2n\\ 0 \end{matrix}\right) +\left(\begin{matrix}2n\\ 2 \end{matrix}\right) +\left(\begin{matrix}2n\\ 4 \end{matrix}\right) + \ldots +\left(\begin{matrix}2n\\ 2k \end{matrix}\right) + \ldots +\left(\begin{matrix}2n\\ 2n \end{matrix}\right)

is equal to

\left(\begin{matrix}2n\\ 1 \end{matrix}\right) +\left(\begin{matrix}2n\\ 3 \end{matrix}\right) +\left(\begin{matrix}2n\\ 5 \end{matrix}\right) + \ldots +\left(\begin{matrix}2n\\ 2k+1 \end{matrix}\right) + \ldots +\left(\begin{matrix}2n\\ 2n-1 \end{matrix}\right) . Can anyone please give feedback and tell me if my solution is correct.

user153017
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  • http://math.stackexchange.com/questions/807423/binomial-expression – lab bhattacharjee May 25 '14 at 06:02
  • Welcome to math.se! For formatting mathematics don't forget that you need dollar signs "$x$" if you want the expression to occur inline, double dollar signs "$$x$$" if you want it on a line by itself - see my edit. Also, please try to choose a readable title ;-) – David May 25 '14 at 06:10

1 Answers1

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Hint: From binomial formula $$(1+x)^{2n}=\sum_{k=0}^{2n}\binom{2n}{k}x^k$$ for $x=-1$

Adi Dani
  • 16,949