Question

Sum to n terms the series $1^2-2^2+3^2-4^2+5^2-6^2+.........$

Answer 1

$S=(1^2-2^2)+(3^2-4^2)+(5^2-6^2)...$
Let $T_1=1^2-2^2$
$T_2=3^2-4^2$
Hence
$T_n=(2n-1{)}^2-(2n{)}^2$
Writing the general term, we get
$(2n-1{)}^2-(2n{)}^2$
$=4n^2-4n+1-4n^2$
$=1-4n$
Hence
$\sum (1-4n)$
$=n-2\left(n\right)(n+1)$
$=n(1-2n-2)$
$=-n(2n+1)$.