# Sum of hypergeometric function

I am trying to evaluate the following sum

$$\sum_{n=1}^N {}_2F_1(-n,n-N,1,x) y^n$$

I notice that according to wolfram alpha, $$\sum_{n=1}^\infty {}_2F_1(-n,b,c,x) y^n = \frac{_2F_1(1,b,c,\frac{yx}{x-1})}{1-y}$$

However, in the question which I want to solve, the b term also varies with n. Any ideas on how to proceed?