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Sum of the series S=1^{2}-2^{2}+3^{2}-4^{2}+......-2008^{2}+20

Question

Sum of the series $S = 1^{2} - 2^{2} + 3^{2} - 4^{2} + ...... - 200 8^{2} + 200 9^{2}$ is

2019045

1005004

2000506

N o n eo f t h ese

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Evelyn Discussed

Sum of the series

S = 1^{2} - 2^{2} + 3^{2} - 4^{2} + ...... - 200 8^{2} + 200 9^{2}

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Text solutionVerified

Solve:

S = 1^{2} - 2^{2} + 3^{2} - 4^{2} + \dots - 200 8^{2} + 200 9^{2}

on adding and substracting the

square of even numbers we get,

\Rightarrow S = (1^{2} + 2^{2} + 3^{2} + 4^{2} + \dots + 200 8^{2} + 2009)^{2} - 2 (2^{4} + 4^{2} + \dots 200 8^{2}) .

\Rightarrow S =

(sum of square of first 2009 natural no.)

- 2 \times (4) \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)]

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

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

= 2019045

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