I am trying to prove by induction that $$\sum_{k=1}^n k^2(k+1) = \frac{1}{12} n(n+1)(n+2)(3n+1)$$
I proved the base case for n=1 easily, however when proving the inductive case for (n+1), I am met with too complex of an expression. I go from $$\sum_{k=1}^{n+1} k^2(k+1) = (\sum_{k=1}^n k^2(k+1)) + (n+1)^2(n+2)$$
After that I distributed the $k^2$ term to get two summations $$(\sum_{k=1}^n k^3) + (\sum_{k=1}^n k^2) + (n+1)^2(n+2)$$
After using the respective equations for $k^3$ and $k^2$ I narrowed it down to $$\frac{1}{4}n^2(n+1)^2 + \frac{1}{6}n(n+1)(2n+1) + (n+1)^2(n+2)$$ and at this point I can't seem to get it to simplify right; I keep getting into messier situations. Do you guys have any tips, or did I do something wrong?
Thank you
Take $n\in\Bbb N$ and assume that$$\sum_{k=1}^nk^2(k+1)=\frac1{12}n(n+1)(n+2)(3n+1).$$Then\begin{align}\sum_{k=1}^{n+1}k^2(k+1)&=\left(\sum_{k=1}^nk^2(k+1)\right)+(n+1)^2(n+2)\\&=\frac1{12}n(n+1)(n+2)(3n+1)+(n+1)^2(n+2)\\&=\frac1{12}(n+1)(n+2)\bigl(n(3n+1)+12n+12\bigr)\\&=\frac1{12}(n+1)(n+2)(3n^2+13n+12).\end{align}On the other hand\begin{align}\frac1{12}(n+1)(n+2)(n+3)(3(n+1)+1)&=\frac1{12}(n+1)(n+2)(n+3)(3n+4)\\&=\frac1{12}(n+1)(n+2)(3n^2+13n+12),\end{align}and therefore$$\sum_{k=1}^{n+1}k^2(k+1)=\frac1{12}(n+1)(n+2)(n+3)(3(n+1)+1).$$