Prove that 1^k+2^k+3^k+...+n^k is devisible by n(n+1)/2 where k is any odd positive integer & n>1 or n=1?

as k is odd we have (a^k+b^k) is divisible by (a+b)

now

n^k + 1 ^k is divisible by (n+1)
(n-1)^k + 2^k is divisible by (n+1)
if n is odd
((n+1)/2)^k is divisible (n+1)/2 so sum is divisible by (n+1)/2

if n is even then there are n/2 terms each is divisible by (n+1)

we consider 2 cases

case 1 n odd :
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the sum is divisible by (n+1)/2

using the same analogy 1^k+2^k+3^k+...+ (n-1)^k is divisible by n(as n-1 is even ) and hence 1^k+2^k+3^k+...+ (n-1)^k + n^k is divisible by n

so it is divisible by n(n+1)/2 as they are co-primes.

case 2 n be even
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the sum is divisible by (n+1)

using the same analogy 1^k+2^k+3^k+...+ (n-1)^k is divsible by n/2 (as n-1 is odd) and hence 1^k+2^k+3^k+...+ (n-1)^k + n^k is divisible by n/2

so it is divisible by n(n+1)/2

so in both cases is is divisible by n(n+1)/2

I am sure some one like will come with a better solution

if n is even we have the sum divisible by n+ 1


if n is odd the middle term is ((n+1)/2)^k

so sum is divisible by (n+1)/2 both cases