112 + 122 + 132 + ... + 202 = ?
(112 + 122 + 132 + ... + 202) = (12 + 22 + 32 + ... + 202) - (12 + 22 + 32 + ... + 102)
Ref: (12 + 22 + 32 + ... + n2) =$\dfrac{1}{6} n(n + 1)(2n + 1)$
= (2870 - 385)
= 2485.
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112 + 122 + 132 + ... + 202 = ? |
Answer |