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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:

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

This implies

$$\mathbf{ 2S = (1^k + n^k ) + (2^k + (n-1)^k ) + \cdots + (n^k + 1^k) }$$
Since k is odd, we know $$\mathbf{ a^k + b^k = (a+b)(a^{k-1} – a^{k-2} b + \cdots + b^{k-1}) }$$ , that is $$\mathbf{ a+b}$$ divides $$\mathbf{a^k + b^k }$$

Applying this to $$\mathbf{ 2S = (1^k + n^k ) + (2^k + (n-1)^k ) + \cdots + (n^k + 1^k) }$$ we have (n+1) divides $$\mathbf{ 1^k + n^k }$$ , $$\mathbf{ (2 + n-1) =(n+1) }$$ divides $$\mathbf{(2^k + (n-1)^k)}$$ and so on. Hence we can take n+1 common from each bracket, leading us to the following expression:

$$\mathbf{ 2S = (n+1)\times(\text{some constant}) \implies S = \frac{(n+1)}{2} \times(\text{some constant}) }$$ . This implies $$\mathbf{ \frac{(n+1)}{2} }$$ divides S if n+1 is even, other wise n+1 divides S.

Now we show n (or n/2) divides S (when is odd or even respectively). To show this we write

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

Again $$\mathbf{ 2S =n^k + (1^k+ (n-1)^k) + (2^k + (n-2)^k) \cdots + ((n-1)^k + 1^k) + n^k }$$
Since k is odd n divides $$\mathbf{ a^k + (n-a)^k }$$ for all a from 1 to n. Hence we can take n as common and have:

$$\mathbf{ 2S = (n)\times(\text{some constant}) \implies S = \frac{(n)}{2} \times(\text{some constant}) }$$ . This implies $$\mathbf{ \frac{(n)}{2} }$$ divides S if n is even, other wise n divides S.

Now gcd of (n, n+1) = 1. Hence $$\mathbf{ \frac{n(n+1)}{2} }$$ divides S.

Proved

May 6, 2014