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Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Class11  >>  Sequence and Series
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Find the sum of the series -1 + 4 - 9 + 16 - 25 +-----+ 4008004.

$\begin{array}{1 1} 410106 \\ 4080408 \\ 401008 \\ 4010006 \end{array}$

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Answer: (d) 4010006
Explanation : The series can be written as $\;((-1)^2+2^2)-3^2+4^2--------+2002^2$
$S_{n}=2^2-1^2+4^2-3^2+-------2002^2-2001^2$
$S_{n}=(2-1)(2+1)+(4-3)(4+3)+--------+(2002-2001)(2002+2001)$
$S_{n}=(1)(1+2+3+4--------+2001+2002)$
$=\frac{2002*2003}{2}\qquad[\sum_{n=1}^N N=\frac{n(n+1)}{2}]$
$=4010006$
answered Dec 31, 2013 by yamini.v
 

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