Let N and K be natural numbers, where k ≥ 2, prove that n ^ k can be written as the sum of N consecutive odd numbers

Let N and K be natural numbers, where k ≥ 2, prove that n ^ k can be written as the sum of N consecutive odd numbers

Let the first odd number be a
Then n ^ k = a + (a + 2) + (a + 4) + [a + 2 (n-1)] = Na + [2 + 4 +... + 2 (n-1)] = Na + n (n-1) = n (a + n-1)
n^(k-1)=a+n-1
a=n^(k-1)-n+1
Because k > = 2, k-1 > = 1, and is a natural number, then n ^ (k-1) - n > 0, so n ^ (k-1) - N + 1 is a natural number
That is to say, as long as we take the first odd number as n ^ (k-1) - N + 1, then starting from it, the sum of consecutive n odd numbers is exactly equal to n ^ K