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sum of the series : 1^2-2^2+3^2-4^2.....+2019^2-2020^2 is

KAUSHIK SUBRAHMANIAN , 8 Years ago
Grade 11
anser 1 Answers
ronit

Last Activity: 8 Years ago

12-22+32-42…..+20192-20202
(1+2)(1-2)+(3+4)(3-4)........+(2019+2020)(2019-2020)
-(3+7+............+4039
using arithmetic progressions,we get -2041210
ANSWER= -2041210

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