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1. (112 + 122 + 132 + ... + 202) = ?





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  • By: guest on 01 Jun 2017 05.46 pm
    (112 + 122 + 132 + ... + 202) = (12 + 22 + 32 + ... + 202) - (12 + 22 + 32 + ... + 102) Ref: (12 + 22 + 32 + ... + n2) = 1 n(n + 1)(2n + 1)     6 =  20 x 21 x 41 - 10 x 11 x 21 6 6 = (2870 - 385) = 2485.
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