Difference between minus one and plus one induction? I recently started a Combinatorics class, in which my teacher (grad student) has instructed us to Prove by induction that $$1^2+2^2+\ldots+n^2 = \frac{n(n+1)(2n+1)}{6} = \frac{2n^3+3n^2+n}{6}$$ this is trivial in the fact that it has been solved many times before, however my professor has insisted I solve it by using $P(n-1)$ as opposed to $P(n+1)$, which I've done below. 
Basis
$$\frac{1(1+1)(2*1+1)}{6} = 1$$
Inductive Step $(n-1)$
$$1^2+2^2+\ldots + (n-1)^2 = \frac{(n-1)(n)(2(n-1)+1)}{6}$$
Which Simplifies to
$$\frac{(n-1)(n)(2n-1)}{6} \rightarrow \frac{2n^3-3n^2+n}{6}$$
Add $6\frac{n^2}{6}$ to both sides and we've proven by induction. 
My question is do there exists any mathematical proofs for which solving by Induction with $n+1$ and $n-1$ are not interchangeable and should I petition my professor to be able to use them interchangeably. I am aware that solving using $n-1$ and $n+1$ is identical, at least for every scenario I've come across (we're working with positive integers so I'm not expecting any variance from that), however given the overwhelming amount of resources, I can't for the life of me figure out why I am being instructed to use a method opposite what seems to be the norm for any other reason besides my teacher's personal preference.

I think the only reason s/he did it this way is s/he wanted to have the final simplification end in the form $\frac {2n^3 + 3n^2 + n}6$
Had he done $P(n) \implies P(n+1)$ it would have involve a lot of factoring to get in the form $\frac {2(n+1)^3 + 3(n+1)^2 + (n+1)}6$.
Try it:
$1 + 2 + .... + n^2 = \frac {2n^3 + 3n^2 + n}6$
So $1 + 2 + ...... + n^2 + (n+1)^2 =  \frac {2n^3 + 3n^2 + n}6 + n^2 + 2n + 1$
$= \frac {2n^3 + 9n^2 + 13n + 6}6$
... and we have to somehow get that to .... $=\frac {2(n+1)^3 + 3(n+1)^2 + (n+1)}6$
... which isn't impossible ...
$\frac {2n^3 + 9n^2 + 13n + 6}6= \frac {2n^3 + 6n^2 + 6n + 2 + 3n^2 + 7n + 4}6$
$=\frac {2(n+1)^3 + 3n^2 + 6n + 3 + n + 1}6$
$= \frac {2(n+1)^3 + 3(n+1)^2 + (n+1)}6$
... but.... why schlep?  That factoring and working backwords isn't the point.
It's easier to follow it and to go to the conclusion if you you work toward simplifying. 
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I suppose an easier compromise would be to do:
$\frac {2(n+1)^2 + 3(n+1)^2 + (n+ 1)}{6} = \frac {2n^3 + 6n^2 + 6n + 2 + 3n^2 + 6n + 3 + n+1}6 = \frac {2n^3 + 9n^3 + 13n + 6}6$
first.
Then:
$1+2 + ..... + n^2 + (n+1)^2 =$
$\frac {2n^3 + 3n^2 + n}6 + (n+1)^2 = \frac {2n^3 + 3n^2 + n}6 +n^2 + 2n + 1=$
$\frac {2n^3 + 3n^2 + n + 6n^2 + 12n + 6}6 = \frac {2n^3 + 9n^3 + 13n + 6}6$
$= \frac {2(n+1)^2 + 3(n+1)^2 + (n+ 1)}{6}$
