Question

If $\sum _{r-1}^{6}l\left(r\right)=n(2n^2+9n+13)$, then find the sum $\sum _{r-1}^6\sqrt{\left(1\right(r)}$

Answer 1

${\sum }_{r=1}^{n}I\left(r\right)=n(2n^2+9n+13)$

$I\left(r\right)=S_{\left(r\right)}-S_{r-1}=S_r-S_{r-1}$

$=r(2r^2+9r+13)-(r-1)\left(2\right(r-1{)}^2+9(r-1)+13)$

$=2r^3+9r^2+13r-(r-1)\left[2\right(r^2-2r+1)+9r-9+13]$

$=2r^3+9r^2+13r-(r-1)[2r^2-4r+2+9r+4]$

$=2r^3+9r^2+13r-(r-1)(2r^2+5r+6)$

$=2r^3+9r^2+13r-2r^3-5r^2-6r+2r^2+5r+6$

$=6r^2+12r+6$

$=6(r^2+2r+1)$

$=6(r+1{)}^2$

$\therefore \sqrt{I\left(r\right)}=\sqrt{6(r+1{)}^2}=\sqrt{6}(r+1)$

$\therefore {\sum }_{r=1}^n\sqrt{I\left(r\right)}={\sum }_{r=1}^n\sqrt{6}(r+1)$

$=\sqrt{6}(\frac{n(n+1)}{2}+n)=\sqrt{6}\left(\frac{n^2+3n}{2}\right)$