Roman C. answered • 05/07/17

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More generally, for odd n,

1

^{n}+ 2^{n}+ 3^{n}+ … + n^{n}is a multiple of n.

To show that, note that n

^{n}is a multiple of n and that we can split the sum into pairs of terms (n-k)^{n}+ k^{n}summing to a multiple of n.We have to expand to see that (n-k)

^{n}+ k^{n}really is divisible by n for odd n:(n-k)

^{n}+ k^{n}= C

_{n,0}n^{n}- C_{n,1}n^{n-1}k + C_{n,2}n^{n-2}k^{2}- … + C_{n,n-1}n k^{n-1}- C_{n,n}k^{n}+ k^{n}= n (C

_{n,0}n^{n-1}- C_{n}_{,1}n^{n-2}k + C_{n,2}n^{n-3}k^{2}- … + C_{n,n-1}k^{n-1})The last equality is justified because C

_{n,n}= 1, so the last two terms canceled.