Prove that $\forall n \in \mathbb{N},$ we can express $n^k$ where $k \geq2$ is an integer, as a sum of $n$ consecutive odd numbers.
My Solution -
Let $2m+1$ be the first odd number , $m \geq 0 , m \in \mathbb{N_0.}$
$\Rightarrow 2m+1 , 2m+3 , 2m+5,...,2m+2n-1 $ are $n$ consecutive odd numbers.
$⇒$ Sum of $n$ odd numbers ,
$⇒\sum_{k=1}^{n} 2m+2k-1 = 2mn+n^2=n(2m+n)$
Now notice that $2m+n$ should be a power of n.
$⇒2m+n=n^p ; $ For some $p\geq2, p \in \mathbb{N}.$
$⇒2m = n^p-n=n(n^{p-1}-1)$
Now $2$ cases :
$\text{Case 1:}$ $n$ is odd $n=2a+1 ; a \geq 0 , a \in \mathbb{N_0}$
$⇒(2a+1)[(2a+1)^{p-1}-1]$
Now note that $(2a+1)^{p-1}-1 \equiv 0 \pmod {2}⇒$ Above expression is divsible 2 $⇒$ Can be expressed in terms of $m$ as $2m$ is even.
$\text{Case 2 : }$ $n$ is even $⇒n=2a ; a \geq 0 , a \in \mathbb{N_0}$
$⇒2a[(2a)^{p-1}-1]$
Now note that $2a[(2a)^{p-1}-1] \equiv 0 \pmod{2}⇒ $ Divisible by $2⇒$ Can be expressed in terms of $m$.
Hence there exists $m$ such that it makes $n^k$.
Hence Proved.
Can someone tell if this proof is mathematically correct?
The following formula describes a sequence of $n$ consecutive odd numbers that sum to $n^k$ for any $n,k$: $$\sum_{i=\frac{n^{k-1}-n}{2}+1}^{\frac{n^{k-1}+n}{2}}(2i-1)$$ The sum contains $n$ members whose average value is $n^{k-1}$; hence the sum is $n\cdot n^{k-1}=n^k$