Question

Find the sum of the series $S=1^2-2^2+3^2-4^2+...+{29}^2-{30}^2$ ?

• A. $-454$
• B. $-464$
• C. $-455$
• D. $-465$

Answer 1

The correct option is D $-465$

$S=1^2-2^2+3^2-4^2+...+{29}^2-{30}^2=(1^2-2^2)+(3^2-4^2)+....+({29}^2-{30}^2)$
Now, $t_n={(2n-1)}^2-{\left(2n\right)}^2=(2n-1+2n)(2n-1-2n)=-(4n-1)$ for $n=1$ to $15$
$S=-\sum _{n=1}^{15}(4n-1)=-4\sum _{n=1}^{15}n+\sum _{n=1}^{15}1=-\frac{-\mathrm{4.15.}(15+1)}{2}+15=-465$
Ans: D