Summation with Binomial Expansion

[Jake1802]

Homework Statement

How can i prove this relationship
\sum _{i=0}^k \text{Binomial}[n+1,k-2i] - \sum _{i=0}^k \text{Binomial}[n,k-2i]=\sum _{i=0}^k \text{Binomial}[n,k-1-2i]

Homework Equations

Binomial (n,k)=n^k/k!

The Attempt at a Solution

I attempted subbing into mathyematica but this didn't work so i attempted by hand and got completely lost. Any helpful comments would be helpful.
Result from Mathematica
-Binomial[n, -1 + k] HypergeometricPFQ[{1, 1/2 - k/2,
1 - k/2}, {1 - k/2 + n/2, 3/2 - k/2 + n/2}, 1] -
Binomial[n,
k] HypergeometricPFQ[{1, 1/2 - k/2, -(k/2)}, {1/2 - k/2 + n/2,
1 - k/2 + n/2}, 1] +
Binomial[1 + n,
k] HypergeometricPFQ[{1, 1/2 - k/2, -(k/2)}, {1 - k/2 + n/2,
3/2 - k/2 + n/2}, 1]

 

[vela]

Fixed your LaTex. Is this the relation you're supposed to prove?

$$\sum_{i=0}^k \begin{pmatrix}n+1\\k-2i\end{pmatrix} - \sum_{i=0}^k \begin{pmatrix}n\\k-2i\end{pmatrix}=\sum_{i=0}^k \begin{pmatrix}n\\k-1-2i\end{pmatrix}$$

I don't think it's correct because k-2i<0 for some values of i in the summation.

Binomial (n,k)=n^k/k!

That's not right. It should be

$$\begin{pmatrix}n\\k\end{pmatrix}=\frac{n!}{k!(n-k)!}$$

 

[Jake1802]

but when k-2i<0 the value will be zero