Q:

Equality of two expressions

Is the following true for even integer n = 2m > 1?
$\sum_{k=1}^m 2^{2k-1} 
\left( {\begin{array}{*{20}c}
n \\
2k-1 \\
\end{array}} \right)
    = \sum_{k=1}^m 2^{2k} 
\left( {\begin{array}{*{20}c}
n \\
2k \\
\end{array}} \right)$

A:

First replace n by 2m on both sides. 
Now we have to prove -
$$
\sum\limits_{k=1}^{m}2^{2k-1}{2m \choose 2k-1} = \sum\limits_{k=1}^{m}2^{2k}{2m \choose 2k}
$$
Now consider the following two equations - 
$$
(x + 1)^{2m} = {2m \choose 0}x^0 + {2m \choose 1}x^1 + {2m \choose 2}x^2 + ... + {2m \choose 0}x^{2m -1} + {2m \choose 2m}x^{2m}
$$
$$
(x - 1)^{2m} = {2m \choose 0}x^0 - {2m \choose 1}x^1 + {2m \choose 2}x^2 - ... - {2m \choose 0}x^{2m -1} + {2m \choose 2m}x^{2m}
$$
Adding them we have -
$$
\frac{(x - 1)^{2m} + (x + 1)^{2m}}{2} = {2m \choose 0}x^0 + {2m \choose 2}x^2 + ...  + {2m \choose 2m}x^{2m} = {2m \choose 0}x^0 + \sum\limits_{k=1}^{m}x^{2k}{2m \choose 2k}
$$
Subtracting them we have -
$$
\frac{(x + 1)^{2m} - (x + 1)^{2m}}{2} = {2m \choose 1}x^1 + {2m \choose 3}x^3 + ...  + {2m-1 \choose 2m}x^{2m-1} = \sum\limits_{k=1}^{m}x^{2k-1}{2m \choose 2k-1}
$$
Now in the 2 equations just derived substitute x = 2. This results in -
$$
\frac{(2-1)^{2m} + (2+1)^{2m}}{2} = {2m \choose 0} + \sum\limits_{k=1}^{m}2^{2k}{2m \choose 2k} =: (a)
$$
$$
\frac{(2+1)^{2m} - (2-1)^{2m}}{2} = \sum\limits_{k=1}^{m}2^{2k-1}{2m \choose 2k-1} =:(b)
$$
Now manipulating (a) we have -
$$
\frac{3^{2m} - 1}{2} = \sum\limits_{k=1}^{m}2^{2k}{2m \choose 2k}
$$
This is equal to the the left hand side of (b).
Hence proved.

