Manipulating a symmetric sum indices

Question

$$ \sum_{}^{} \binom{2n+1}{i}( 6 \sqrt{6})^i 14^{2n+1-i}$$

Over the sequence,

$$ i= \{1,3,5,..,2n+1 \}$$

Suppose I do the substitution, $ 2n+1 - i = j$,

$$ \sum_{j}^{} \binom{2n+1}{j} ( 6 \sqrt{6})^{2n+1 -j} 14^{j}$$

Note that the value of $j$ we are summing over is same here but it turns out in this new sum, the power of $ (6 \sqrt{6} )$ is even..? (Note: that $2n+1-j$ is even because $2n+1$ and $j$ are odd.)

Answer 1

The answer is: $2n + 1 - j$ is odd. Because $j$ is even.

You can check writing for $n = 1$.
The first sum: $$ \binom{3}{1}(6\sqrt{6})^1(14)^2 + \binom{3}{1}(6\sqrt{6})^3(14)^0 $$ Then, the second sum ... $$ \binom{3}{2}(6\sqrt{6})^{3 - 2}(14)^2 + \binom{3}{0}(6\sqrt{6})^{3 - 0}(14)^0 $$