For this question, I'm stuck on the inductive step for this proof. Here is what I have so far. Can anyone please help me out? Thanks
$\sum_{i=0}^n i = \frac{n(n+1)}{2}$.Use this result to prove that if m and n are any positive integers and m is odd, then $\sum_{i=0}^{m-1} (n+k) $ is divisible by m. Does the conclusion hold if m is even? Justify your answer
Base Case: m = 3, n = 0
$\frac{3}{(0+1+2)}$
$\frac{3}{3}$
Let r be an odd integer such that m = 2r+1
Inductive Step:
2r+1|$\sum_{i=0}^{2r} (n+k) $ -> 2r+2|$\sum_{i=0}^{2r} (n+k) $
I will assume that you want to show that $\sum_{k=0}^{m-1} (n+k) $ is divisible by $m$.
Then
$\begin{array}\\ s(m, n) &=\sum_{k=0}^{m-1} (n+k)\\ &=\sum_{k=0}^{m-1} n+\sum_{k=0}^{m-1} k\\ &=mn+\dfrac{(m-1)m}{2}\\ \end{array} $
If $m$ is odd then $m=2j+1$ for some integer $j$ so
$\begin{array}\\ s(m, n) &=mn+\dfrac{(2j+1-1)m}{2}\\ &=mn+\dfrac{(2j)m}{2}\\ &=mn+jm\\ &=m(n+j)\\ \end{array} $
is divisible by $m$.