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Q)

Find the sum of the series $1^2$ - $2^2$ + $3^2$ - $4^2$ + $5^2$ - $6^2$.....$1000^2$.

$(a)\;505050\qquad(b)\;500050\qquad(c)\;50005\qquad(d)\;-500500$

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A)
Answer : (d) -500500
Explanation : $S_{n}=1^2-2^2+3^2-4^2--------------------1000^2$
$S_{n}=(1^2-2^2)+(3^2-4^2)+------------(999^2-1000^2)$
$=(1-2)(1+2)+(3-4)(3+4)+-----------------+(999-1000)(999+1000)$
$=(-1)(1+2)+(-1)(3+4)+-----------------(-1)(999+1000)$
$=(-1)[1+2+3+4+-----------------999+1000]$
$=(-1)\frac{1000*1001}{2}\qquad[\sum_{n=1}^{N} N=\frac{n(n+1)}{2}]$
=-500500
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