How to solve binomial sum but with denominator

Question

Simple question. How does one give a closed form for this ugly boy? $$\sum_{i=1}^{k-1}\binom{k-1}{i}\frac{(1+p)^{k-1-i}(2x-1-p)^{i}}{i}$$

Answer 1

As Donald Splutterwit commented, there is a problem if the summation starts at $i=1$.

However, for $$S_k=\sum_{\color{red}{i=2}}^{k-1}\binom{k-1}{i}\frac{(1+p)^{k-1-i}(2x-1-p)^{i}}{i-1}$$ let $t=\frac{2 x-p-1}{p+1}$ to get $$S_k=(p+1)^{k-1}\sum_{{i=2}}^{k-1} \binom{k-1}{i}\frac{t^i }{i-1}$$ which leads to $$S_k=\frac{1}{2} (k-2) (k-1) (p+1)^{k-1}\,t^2\,\, _3F_2(1,1,3-k;2,3;-t)$$ where appears an hypergeometric function (which is still a summation).