<s>[INST] How to rewrite $2n^3+9n^2+13n+6$ as a polynomial in terms of $n+1$? I'm working on an induction proof for proving that the sum of the squares of the first n natural numbers is equal to: $$\frac{2n^3 +3n^2 + n}6$$  I understand how to get to the answer starting in the form of: $$\frac{n(n+1)(2n+1)}6$$ which just seems to require simpler manipulations.
I've solved for the series of: $$1^2 + 2^2 + \space...\space+ n^2 +(n+1)^2 = \frac{2n^3+9n^2+13n+6}6$$
I know that using synthetic division I get the factors as $(n+1), (n+2), (2n+3)$ or $$\frac{(n+1)*(n+2)*(2n+3)}6$$ I just can't figure out how to manipulate the factors from here to give me: $$\frac{2(n+1)^3+3(n+1)^2+(n+1)}6$$
How would I manipulate the equation to get it in terms of $(n+1)$ ?
 [/INST] $(n + 1)(n + 2)(2n + 3) = (n + 1)((n + 1) + 1)(2(n + 1) + 1) = ((n + 1)^2 + (n+ 1))(2(n + 1) + 1) = 2((n + 1)^2 + (n + 1))(n + 1) + (n + 1)^2 + (n + 1)= 2(n + 1)^3 + 3(n + 1)^2 + (n + 1)$
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