Are these two summation expressions equal to each other?

Question

Is this equality true?

$$n!\sum_{k=0}^n \left(\frac{1^k + (-1)^k}{2(\frac{k}{2}!)} \right).\left( \frac{1}{(n-k)!} \right) = \sum_{k=0}^{\lfloor\frac{n}{2}\rfloor}\frac{(2k)!}{k!}.\binom{n}{2k}$$

Answer 1

Since the left side is nonzero only when $k$ is even, let $k=2j$ to get

$\displaystyle n!\sum_{k=0}^n \left(\frac{1^k + (-1)^k}{2(\frac{k}{2}!)} \right).\left( \frac{1}{(n-k)!} \right)=n!\sum_{j=0}^{\lfloor\frac{n}{2}\rfloor}\frac{1}{j!}\cdot\frac{1}{(n-2j)!}=\sum_{j=0}^{\lfloor\frac{n}{2}\rfloor}\frac{n!}{j!(n-2j)!}$

$\displaystyle=\sum_{j=0}^{\lfloor\frac{n}{2}\rfloor}\frac{(2j)!}{j!}\cdot\frac{n!}{(2j)!(n-2j)!}=\sum_{j=0}^{\lfloor\frac{n}{2}\rfloor}\frac{(2j)!}{j!}\binom{n}{2j}$

Answer 2

On the right you have $\sum_{k=0}^{\lfloor n/2\rfloor} \frac {n!}{k!(n-2k)!}$.

On the left, notice that $\frac{1^k + (-1)^k}{2}$ is 1 when $k$ is even and 0 when $k$ is odd. Changing indexes carefully gives the equality.

Answer 3

To expand on justt's answer, note that the LHS is $$ \sum_{k=0}^n \frac{1+(-1)^k}{2} {n \choose k} \frac{k!}{(k/2)!}. $$ [Also, what do you mean by $(3/2)!$, ($k=3$)?]