Something something Riemann sum to integral is the most that I remember. I just don't remember how we did it or whether or not that would be the best method for doing it.
Let $ \theta(n) = \sum\limits_{k=1}^{n} k $
and $ \phi(n) = \sum\limits_{k=1}^{n} 2k-1 $
Let $ P(n) $ be the statement $ n^3 = \psi(n)$, where $ \psi(n) = \phi(\theta(n)) - \phi(\theta(n-1)) $.
I need to prove that $ P(n) $ is true for all positive $ n $ by induction. My only progress was defining the series provided in terms of the functions that I created, $ \theta(n) $ and $ \phi(n) $. To complete the proof, I need those two functions in algebraic form. Can anybody point me in the right direction?
You have
\begin{align} \sum_{k=1}^{n} k &= 1 + 2 + 3 + \cdots + n \\ &= n + (n - 1) + (n - 2) + \cdots + 1 \\ \end{align}
where in the second line we just reversed the order in which we summed. Now if we add the two lines, notice that term by term we get $n + 1$, i.e
\begin{align} (n) + 1 &= n + 1 \\ (n - 1) + 2 &= n + 1 \\ (n - 2) + 3 &= n + 1 \\ \vdots \end{align}
and hence we find
\begin{align} 2 \sum_{k=1}^{n} k &= \underbrace{(n + 1) + (n + 1) + \cdots + (n + 1)}_{\text{n times}} \\ &= n(n + 1) \\\\ \implies \sum_{k=1}^{n} k &= \frac{1}{2} n(n + 1) \end{align}
Now try it for the other series, you should get
$$\sum_{k=1}^{n} (2k - 1) = n^{2}$$