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.
Let s(k)=sum of kpowers of first n digits.
s(1)=n(n+1)/2=a.
s(5)=(4a^3a^2)/3.
s(7)=(6a^44a^3+a^2)/3.
So s(5)+s(7)=2a^4 as desired.
Proof of each formula follows by induction. You could use the calculus of differences to derive each, if you don't mind the tedium and you don't make a mistake, but it's easier and lots more fun to use the net.
Pascal got there early.
Bernoulli numbers play a part.
The end(?) is Faulhaber's Fabulous Formula.

Posted by xdog
on 20171210 12:30:44 