Question

4²+7²+...+(3n+1)²=S, what is S?

Answer 1

`\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((n(n+1)(2n+1))/(6)\right)+6\left((n(n+1))/(2)\right)+n=\\\\(3n(n+1)(2n+1))/(2)+3n(n+1)+n=\\\\(3n(2n^2+n+2n+1))/(2)+3n^2+3n+n=\\\\(3n(2n^2+3n+1))/(2)+3n^2+4n=\\\\(6n^3+9n^2+3n)/(2)+(6n^2+8n)/(2)=\\\\(6n^3+15n^2+11n)/(2)`