I came across this question while doing $\text{6.042J}$ from MITOCW. I have a doubt in the part c, namely, why do we need to manipulate the formula in that way?
Here is my solution so far to the other parts: Let us assume $\text{C}$ is the set of counter-examples for which the formula isn't true.
$\text{(a)}$ There must be a least $m$ as $n \in \mathbb{Z^+}$, which is the set of positive integers (given, $n > 0$). So, by the Well-Ordering Principle, there must exist a least $m$
$\text{(b)}$ Clearly, $m \neq 1$ as for $n = 1$, we have, $\sum_{i=0}^{0}(2i+1)=2(0)+1=1=n^2$, so the formula holds true. So, clearly, $m \geq 2$
Now, here is my doubt. For part $\text{(c)}$, what I think they have done, is replaced $i$ by $i-1$. Am I right here? Because I see that replacing $i$ by $i-1$, we have to consequently change the lower limit to $i=1$ and the upper limit to $n$, so the formula would look like,
$$\sum_{i=1}^{n}(2(i-1)+1)=n^2$$ Is this correct? And then as $m$ is the least element such that the formula doesn't hold, the formula must hold for $m-1\geq 1 \text{from part (b)}$, and then we replace $n$ by $m-1$ in the above formula to obtain $(2.10)$. Is this reasoning okay? But what I still am confused by is why they had to change the limits of the sum to show this? Or is it just an exercise in changing the limits?
It's not quite as simple as replacing a variable with another. Note that, taking $n := m-1$ in $(2.9)$,
$$\sum_{i=0}^{(m-1)-1} (2i+1) = (m-1)^2$$
is effectively our starting point. (We know this much is true since $m$ is presumed our "first failure", or rather $m-1$ is our last-ensured success.)
We can reindex by letting $k := i+1$. Then:
We get
$$\sum_{i=0}^{(m-1)-1} (2i+1) = \sum_{k=1}^{m-1} (2(k-1)+1) = (m-1)^2$$
In short, nothing really deep going on here (other than the validity of the claim), just algebra.