Proving $\sum_{k=0}^n\frac{3^{k+4}\binom{n}{k}}{\binom{k+4}4}+\sum_{m=0}^3\frac{\binom{n+4}{m}3^m}{\binom{n+4}4}=\frac{4^{n+4}}{\binom{n+4}4}$

Question

Prove that $$\sum_{k=0}^n\frac{3^{k+4}\binom{n}{k}}{\binom{k+4}{4}} +\sum_{m=0}^3\frac{\binom{n+4}{m}3^m}{\binom{n+4}{4}} =\frac{4^{n+4}}{\binom{n+4}{4}}$$

Wolfram Alpha shows that both expressions are equal but I can't prove it mathematically.

I am not able to calculate each series individually let alone the whole expression.

Any help is greatly appreciated.

Answer 1

As Bruno B suggested in a comment, multiply by $\binom{n+4}4$ and simplify the terms in the first sum:

$$ \binom{n+4}4\frac{\binom nk}{\binom{k+4}4}=\frac{(n+4)!}{n!4!}\frac{\frac{n!}{k!(n-k)!}}{\frac{(k+4)!}{k!4!}}=\frac{(n+4)!}{(k+4)!(n-k)!}=\binom{n+4}{k+4}\;. $$

Then merge the two sums, with $m=k+4$, to get the sum for the binomial expansion of $(3+1)^{n+4}$.