<s>[INST] Proof by induction (Combinatorics) This is my first question on here so please bear with me, thank you.

Prove that for all positive integers $n$
  $$2(1+2+...+n)^4 = (1^5 + 2^5 +...+ n^5) + (1^7 + 2^7 +...+ n^7)$$

After establishing the base case, I proceeded to the induction:
$2(1+2+...+(n+1))^4 = (1^5 + 2^5 +...+ (n+1)^5) + (1^7 + 2^7 +...+ (n+1)^7)$
$2(1+2+...+(n+1))^4 = (1^5 + 2^5 +...+ n^5)+(n+1)^5 + (1^7 + 2^7 +...+n^7)+(n+1)^7$
$2(1+2+...+(n+1))^4 = 2(1+2+...+n)^4 + (n+1)^5 + (n+1)^7$
$2(1+2+...+(n+1))^4 - 2(1+2+...+n)^4 = (n+1)^5 + (n+1)^7$
$2[(1+2+...+(n+1))^4 - (1+2+...+n)^4] = (n+1)^5 + (n+1)^7$
And from here I've hit a wall and can't figure out how to continue. I tried substituting $(1+...+n) = [n(n+1)]/2$ and $(n+1)=[2(1+...+n)]/n$ but to no avail.
 [/INST] You must prove $$2\left(\sum_{i=1}^{n+1}i\right)^4 = \sum_{i=1}^{n+1}i^5 + \sum_{i=1}^{n+1}i^7.$$
Let $S_n = \sum_{i=1}^{n}i = \frac{n(n+1)}{2}$
You have $$2\left(\sum_{i=1}^{n+1}i\right)^4 = 2\left(\left(\sum_{i=1}^{n}i\right)^4 + 4(n+1)S_n^3 + 6(n+1)^2S_N^2 + 4(n+1)^3S_n + (n+1)^4 \right).$$
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