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 $(n+1)$ ?
 A: Say $p(n)=2n^3+9n^2+13n+6$. Taylor expanding this polynomial around $n+1$ gives $$p(n)=p(n+1)-p'(n+1)+\frac{1}{2}p''(n+1)-\frac{1}{6}p'''(n+1)$$
The right-hand side is easy to calculate, giving you an equivalent polynomial in powers of $n+1$. You can see immediately that the coefficients of the third-degree terms must both be 2, because the derivatives kill the third degree. By similar inspection (instead of unnecessary calculation), noting which powers are killed by the derivatives, the coefficient of the second-degree term will be $9-6=3$, the coefficient of the first-degree term will be $13-18+6=1$, and the constant will be $6-13+9-2=0$. This gives 
$$p(n)=2(n+1)^3+3(n+1)^2+(n+1)$$
as desired. Note that there is no need to factor $p(n)$, as you did.
A: replace $n$ with $((n+1)-1)$ or $(m-1)$ (where $m$ is your new variable $n+1$) and expand.
A: $(n + 1)(n + 2)(2n + 3) = (n + 1)((n + 1) + 1)(2(n + 1) + 1) = ((n + 1)^2 + (n+ 1))(2(n + 1) + 1) = 2((n + 1)^2 + (n + 1))(n + 1) + (n + 1)^2 + (n + 1)= 2(n + 1)^3 + 3(n + 1)^2 + (n + 1)$
A: Let $a_3(n+1)^3 + a_2(n+1)^2+ a_1(n+1) +a_0 = 2n^3+9n^2+13n+6 $.
Now expand the LHS and equate the coefficients of $n^3, n^2 ,n$ and constant to $2,9,13,6$ respectively and solve to find $a_0,a_1,a_2,a_3$.
