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Question

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


Options:
A .  385
B .  2485
C .  2870
D .  3255
Answer: Option B

(112 + 122 + 132 + ... + 202) = (12 + 22 + 32 + ... + 202) - (12 + 22 + 32 + ... + 102)

(112 + 122 + 132 + ... + 202) = ? Ref: (12 + 22 + 32 + ... + n2) = 1 n(n + 1)(2n + 1) (112 + 122 + 132 + ... + 202) = ?     6

= (2870 - 385)

= 2485.



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