<s>[INST] odd/even binomial coefficient identity 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.
 [/INST] Hint:
From binomial formula $$(1+x)^{2n}=\sum_{k=0}^{2n}\binom{2n}{k}x^k$$ for $x=-1$
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