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Sum of odd powers of n consecutive numbers (TOMATO subj 31)

If k is an odd positive integer, prove that for any integer \(\mathbf{ n ge 1 , 1^k + 2^k + \cdots + n^k }\) is divisible by \(\mathbf{ \frac {n(n+1)}{2} }\)

Discussion:
We write the given expression in two ways:

\(\mathbf{ S= 1^k + 2^k + \cdots + n^k }\)
\(\mathbf{ S =n^k + (n-1)^k + \cdots + 1^k } \)

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May 6, 2014

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