# Binomial sum for an arbitrary function

I'm looking for some known results for sum of this type but I can't find anything. The sum is defined as: $$S(x,a,b,n)=\sum_{k=0}^n \binom{n}{k} (-1)^{k} f((a(n-k)+bk)x)$$ where $$f$$ is an arbitrary function and $$a$$ and $$b$$ are real and positive constants. For example for the function $$f(x)=e^{x}$$ the sum can be evaluated with Binomial Theorem and we have: $$S(x,a,b,n)=(e^{ax}-e^{bx})^{n}$$ But for arbitrary function $$f$$? Thanks.

• I suspect there is no compact representation of the sum without knowledge of $f$. – David G. Stork Dec 19 '18 at 16:43
• There probably isn't a single method for evaluating the sum that works for all functions. If you happen to know a little bit about $f$, you can make certain generalizations, though. For example, if $f$ is linear, then $f((a(n-k)+bk)x)=f(anx)-f(akx)+f(bkx)$. – R. Burton Dec 19 '18 at 17:35
• Thanks!! Yes, for an arbitrary function isn't improbable to obtain a closed form. I tried to obtain an Integral representation of the sum via Integral Transform but with no great results. – Papemax89 Dec 19 '18 at 17:40