Summation up to $n$ terms : $\sum r\cdot (r+1)^2$ Summation up to $n$ terms : $$\sum_{r=1}^{n} r\cdot (r+1)^2$$
My attempt :
$$\sum_{r=1}^{n} r\cdot (r+1)^2=\sum_{r=1}^{n} r^3+2\sum_{r=1}^{n} r^2+\sum_{r=1}^{n} r$$
$$\sum_{r=1}^{n} r\cdot (r+1)^2=\sum_{r=1}^{n} r^3+2\sum_{r=1}^{n} r^2+\sum_{r=1}^{n} r$$
But for this method I should know the answers of $\sum r^3$ , $\sum r^2$ and $\sum r$.
Is there another method to find the sum without using $\sum r^3$ , $\sum r^2$ and $\sum r$ ?

Let $f(r)=a+br+cr^2+dr^3+er^4$.
Then $$f(r+1)-f(r)$$
$$=a+b(r+1)+c(r^2+2r+1)+d(r^3+3r^2+3r+1)+e(r^4+4r^3+6e^2+4e+1)-(a+br+cr^2+dr^3+er^4)$$
$$=b+2cr+c+3dr^2+3dr+d+4er^3+6er^2+4er+e$$
$$=(b+c+d+e)+(2c+3b+4e)r+(3d+6e)r^2+4er^3$$
We need this to equal $r(r^2+2r+1)=r+2r^2+r^3$.
So $4e=1$ etc...
