Is $\sum_{x=1}^n (3x^2+x+1) = n^3+2n^2+3n$? I wanna check if the following equation involving a sum is true or false? How do I solve this? Please help me.
$$ \sum_{x=1}^n (3x^2+x+1) = n^3+2n^2+3n$$
for all $n \in \{0,1,2,3, \dots\}$.

$(x+1)^3=x^3+3x^2+3x+1\iff3x^2+3x+1$ is the difference of two consecutive cubes, so the sum will be telescopic, since $3x^2+x+1=(3x^2+3x+1)-2x=\Big[(x+1)^3-x^3\Big]-2x$. $$\sum_{x=1}^n(3x^2+x+1)=\sum_{x=1}^n\Big[(x+1)^3-x^3\Big]-2\sum_{x=1}^nx=\Big[(n+1)^3-1^3\Big]-2\bigg[\frac{n(n+1)}2\bigg]=$$ $$=\Big[(n^3+3n^2+3n+1)-1\Big]-(n^2+n)=n^3+2n^2+2n.$$ See Faulhaber's formula for more information.
