Consider:
(1^{5} + 2^{5}) + (1^{7} + 2^{7}) = 2 *(1 + 2)^{4}
(1^{5} + 2^{5} + 3^{5}) + (1^{7} + 2^{7} + 3^{7}) = 2 *(1 + 2 + 3)^{4}
(1^{5} + 2^{5} + 3^{5} + 4^{5}) + (1^{7} + 2^{7} + 3^{7} + 4^{7}) = 2 *(1 + 2 + 3 + 4)^{4}
... ... and so on ...
First, verify that both sides are equal for further increase in n,
then prove it.
Start with RHS = 2(n/2 (n + 1))^4 = 1/8 (n^2 + n)^4
Then LHS, considering each bracket separately:
Sum (1 to n) n^5 = 1/6 (n^2 + n)^3  1/12 (n^2 + n)^2
Sum (1 to n) n^7 = 1/8 (n^2 + n)^4  1/6 (n^2 + n)^3 + 1/12 (n^2 + n)^2
So the n^5 term exactly cancels out of the n^7 term, leaving 1/8 (n^2 + n)^4, as was to be shown. Very neat.

Posted by broll
on 20171210 21:33:38 