Consider:
(15 + 25) + (17 + 27) = 2 *(1 + 2)4
(15 + 25 + 35) + (17 + 27 + 37) = 2 *(1 + 2 + 3)4
(15 + 25 + 35 + 45) + (17 + 27 + 37 + 47) = 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 k-powers of first n digits.
s(1)=n(n+1)/2=a.
s(5)=(4a^3-a^2)/3.
s(7)=(6a^4-4a^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.
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Posted by xdog
on 2017-12-10 12:30:44 |
statements and references
