Evaluate 1^2-2^2 +3^2 - 4^2 + ...-2008^2+2009^2
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Answer:
Solve:
S=1
2
−2
2
+3
2
−4
2
+…−2008
2
+2009
2
on adding and substracting the
square of even numbers we get,
⇒S=(1
2
+2
2
+3
2
+4
2
+…+2008
2
+2009)
2
−2(2
4
+4
2
+…2008
2
).
⇒S= (sum of square of first 2009 natural no.)
−2×(4)×( sum of square of first 1004
natural no. )
⇒S=
6
2009×2010×4019
−2⋅2
2
(1+2
2
+3
2
+…+(1004)
2
)
⇒S=
6
2009
[2010×4019−8(1004×1005)]
⇒S=
6
2009
×2010[4019−4016]
⇒S=
6
2009
×2010×3=2009×1005
=2019045
Explanation:
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