[tex]\displaystyle\\4^2+7^2+\ldots+(3n+1)^2=\\\\\sum_{k=1}^n(3k+1)^2=\\\\\sum_{k=1}^n(9k^2+6k+1)=\\\\\sum_{k=1}^n9k^2+\sum_{k=1}^n 6k+\sum_{k=1}^n1=\\\\9\sum_{k=1}^nk^2+6\sum_{k=1}^n k+n=\\\\9\left(\dfrac{n(n+1)(2n+1)}{6}\right)+6\left(\dfrac{n(n+1)}{2}\right)+n=\\\\\dfrac{3n(n+1)(2n+1)}{2}+3n(n+1)+n=\\\\\dfrac{3n(2n^2+n+2n+1)}{2}+3n^2+3n+n=\\\\\dfrac{3n(2n^2+3n+1)}{2}+3n^2+4n=\\\\\dfrac{6n^3+9n^2+3n}{2}+\dfrac{6n^2+8n}{2}=\\\\\dfrac{6n^3+15n^2+11n}{2}[/tex]
