General method to find sum of binomial

Question

Specifically I want to ask the method to solve $$\sum_{k=0}^n \binom{4n+b}{4k};\ b=[0,1,2,3]$$

And how to solve series of type $$\sum_{k=0}^n \binom{an+b}{ak};\ a=[1,2,3,...],\ b=[0,1,2,...,a-1]$$

Answer 1

I am sure that you will enjoy the result !

If $$S_n=\sum_{k=0}^n \binom{4n+b}{4k}$$ then a CAS gives

$$4S_n=2^{\frac{4 n+b}{2} } \left((-1)^b\, 2^{\frac{4 n+b}{2} } \cos (\pi b)+\cos \left(\frac{ (4 n+b)\pi}{4} \right)+(-1)^b \cos \left(\frac{3(4 n+b)\pi}{4} \right)\right)-$$ $$4 \binom{4 n+b}{4 n+4} \, _5F_4\left(1,\frac{4-b}{4},\frac{5-b}{4},\frac{6-b}{4},\frac {7-b}{4};\frac{4n+5}{4},\frac{4n+6}{4},\frac{4n+7}{4},\frac{4n+8}{4};1\right)$$ where appears the generalized hypergeometric function.

For the first values of $b$, the sequences can be found in $OEIS$.

$$\left( \begin{array}{cc} b & \text{OEIS sequence} \\ 0 & \text{A070775} \\ 1 & \text{A090407} \\ 2 & \text{A001025} \\ 3 & \text{A090408} \end{array} \right)$$ Be sure that I did not make this by myself.