Question

Sum of the series $S=1^2-2^2+3^2-4^2+......-{2008}^2+{2009}^2$ is

• A. $2019045$
• B. $1005004$
• C. $2000506$
• D. $Noneofthese$`

Answer 1

Answer: (A)

$2019045$

$\text{Solve:}$

$S=1^2-2^2+3^2-4^2+\dots -{2008}^2+{2009}^2$

$\text{on adding and substracting the}$

$\text{square of even numbers we get,}$

$\Rightarrow S={(1^2+2^2+3^2+4^2+\dots +{2008}^2+2009)}^2-2(2^4+4^2+\dots {2008}^2).$

$\Rightarrow S=$ (sum of square of first 2009 natural no.)

$-2\times \left(4\right)\times ($ sum of square of first 1004

natural no. $)$

$\Rightarrow S=\frac{2009\times 2010\times 4019}{6}-2\cdot 2^2(1+2^2+3^2+\dots +(1004{)}^2)$

$\Rightarrow S=\frac{2009}{6}[2010\times 4019-8(1004\times 1005\left)\right]$

$\Rightarrow S=\frac{2009}{6}\times 2010[4019-4016]$

$\Rightarrow S=\frac{2009}{6}\times 2010\times 3=2009\times 1005$

$=2019045$