Problem to prove for all even integers

Question

Prove that $5^n = 3^n + \dfrac{16n (3^{n-2})}{2} + \dfrac{256n (n-2) 3^{n-4}}{8}+ ....+ 4^n$ for all even integers.

I tried finding a pattern, but was unable to do so. Any help would be appreciated.

Answer 1

It is just Binomial Theorem. Let $n=2k$. $$5^{2k}=(3^2+4^2)^k=(3^2)^k+{k\choose 1}(4^2)(3^2)^{k-1}+{k\choose 2}(4^2)^2(3^2)^{k-2}+\ldots+(4^2)^{k}$$ $$=3^n+16\dfrac n23^{n-2}+256\dfrac {\dfrac n2\left(\dfrac{n}2-1\right)}23^{n-2}+\ldots+4^n$$