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.
Please give me feedback on my answer to this question. Question: For all $ n\geq1:\binom{2n}{0}+\binom{2n}{2}+\binom{2n}{4}+\cdots+\binom{2n}{2k}+\cdots+\binom{2n}{2n} $   is equal to $ \binom{2n}{1}+\binom{2n}{3}+\binom{2n}{5}+\cdots+\binom{2n}{2k+1}+\cdots+\binom{2n}{2n-1} $. Answer: False, since;  Let $a=1,  b=-1$  and $n=1$  ,$k=0$   , using Binomial expression.  $$(a+b)^n=  \sum   \binom{n}{k}  a^{k}b^{n-k}$$  Then $(1+-1)^{2(1)}=\binom{2}{0}1^{0}(-1)^{2-0}+\binom{2}{1}1^{1}(-1)^{2-1}+\binom{2}{2}1^{2}(-1)^{2-2}=\binom{2}{0}-\binom{2}{1}+\binom{2}{2}$  Thus :For all $n\geq1:  \binom{2n}{0}+\binom{2n}{2}+\binom{2n}{4}+\cdots+\binom{2n}{2k}+\cdots+\binom{2n}{2n}   \ne \binom{2n}{1}+\binom{2n}{3}+\binom{2n}{5}+\cdots+\binom{2n}{2k+1}+\cdots+\binom{2n}{2n-1}$  .
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