# Closed form for $\sum_{k=0}^{m} {\binom {m}{k}} a^{k} (b+ck)^N$

Is there a closed form for the following? $$\sum_{k=0}^{m} {\binom {m}{k}} a^{k} (b+ck)^N$$

how about a pretty limit for large $b$.

I have tried using the binomial expansion for the $(b+ck)^N$ to convert the above expression to double sum and then tried to simplify the double sum which was not successful. Any idea if I have find a closed form for it?

• Do you mean ${m\choose k}$? – carmichael561 Jun 28 '16 at 21:35
• yes, thanks, I corrected it now. – Susan_Math123 Jun 28 '16 at 21:36

Define $$f(x) = \sum_{k=0}^{m} {\binom {m}{k}} x^{b+ck} = x^b (1+x^c)^m.$$ Define the differential operator $D(g(x)) = xg'(x)$, so that $D^2g(x)=D(D(g(x)) = xg'(x) + x^2g''(x)$, etc. Then $$D^N(f(x)) = \sum_{k=0}^{m} {\binom {m}{k}} D^N(x^{b+ck}) = \sum_{k=0}^{m} {\binom {m}{k}} (b+ck)^Nx^{b+ck},$$ and so $$\sum_{k=0}^{m} {\binom {m}{k}} (b+ck)^Na^k = a^{-b/c} (D^Nf)(a^{1/c}).$$ This will allow you to get a closed form for any particular value of $N$, just by applying $D^N$ to $x^b (1+x^c)^m$ and plugging in $x=a^{1/c}$.