Mathematics - Sequence & Series Question with Solution | TestHub

Given,

$2·2^2-3^2+2·4^2-5^2+2·6^2-............$$20$ terms,

$=\sum _{r=1}^{10}\left(2·(2r{)}^2-(2r+1{)}^2\right)$

$=\sum _{r=1}^{10}\left(8r^2-4r^2-4r-1\right)$

$=\sum _{r=1}^{10}\left(4r^2-4r-1\right)$

$=4\sum _{r=1}^{10}{r}^2-4\sum _{r=1}^{10}r-\sum _{r=1}^{10}1$

$=\frac{4·10·11·21}{6}-4\frac{10·11}{2}-10$

$=44·35-220-10$$=1540-230=1310$