## zmudz one year ago The closed form sum of $$12 \left[ 1^2 \cdot 2 + 2^2 \cdot 3 + \ldots + n^2 (n+1) \right]$$ for $$n \geq 1$$ is $$n(n+1)(n+2)(an+b).$$ Find $$an + b$$.

1. anonymous

You can use equations (22) and (23) from here: http://mathworld.wolfram.com/PowerSum.html

2. anonymous

$$S=12\sum_{k=1}^n k^2(k+1)=12\left(\sum_{k=1}^n k^3+\sum_{k=1}^n k^2\right)=12\cdot\left(\frac{n^2(n+1)^2}4+\frac{n(n+1)(2n+1)}6\right)$$

3. anonymous

so$$n(n+1)\left(3n(n+1)+2(2n+1)\right)$$

4. anonymous

now consider $$3n(n+1)+2(2n+1)=3n^2+7n+2=3n^2+6n+n+2=(n+2)(3n+1)$$