Question

Sum of the series $S=1^2-2^2+3^2-4^2+.....-{2002}^2+{2003}^2$ is

• A. $2007006$
• B. $1005004$
• C. $2000506$
• D. $1005040$

Answer 1

Answer: (A)

$2007006$

$S=1^2-2^2+3^2-4^2+......-{2002}^2+{2003}^2$

$\Rightarrow S=\sum _{k=1}^{2003}{(-1)}^{k+1}{k}^2$

$\Rightarrow S=\sum _{k=1}^{1002}{(2k-1)}^2-\sum _{k=1}^{1001}{\left(2k\right)}^2$

$\Rightarrow S={[(2\times 1002)-1]}^2+\sum _{k=1}^{1001}[{\left(2k\right)}^2-{\left(2k\right)}^2-4k+1]$

$\Rightarrow S={\left(2003\right)}^2-\sum _{k=1}^{1001}(4k-1)={\left(2003\right)}^2-4\times \frac{1001(1001+1)}{2}+1001=2007006$