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Given that 1^2 + 2^2 + 3^2 + … + n^2 = n(n+1)(2n+1)/6, wha

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Given that 1^2 + 2^2 + 3^2 + … + n^2 = n(n+1)(2n+1)/6, wha

by Max@Math Revolution » Tue Apr 16, 2019 12:12 am

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[GMAT math practice question]

Given that 1^2 + 2^2 + 3^2 + ... + n^2 = n(n+1)(2n+1)/6, what is the value of 11^2(1-1/11)(1+1/11) + 12^2(1-1/12)(1+1/12) + 13^2(1-1/13)(1+1/13) + ... + 20^2(1-1/20)(1+1/20)?

A. 1080
B. 2405
C. 2475
D. 2880
E. 3600
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by Max@Math Revolution » Wed Apr 17, 2019 11:52 pm
=>

11^2(1-1/11)(1+1/11) + 12^2(1-1/12)(1+1/12) + 13^2(1-1/13)(1+1/13) + ... + 20^2(1-1/20)(1+1/20)
= 11^2(1-1/11^2) + 12^2(1-1/12^2) + 13^2(1-1/13^2) + ... + 20^2(1-1/20^2)
= (11^2-1) + (12^2-1) + (13^2-1) + ... + (20^2-1)
= (11^2 + 12^2 + 13^2 + ... + 20^2) - 10
= (1^2 + 2^2 + 3^2 + ... + 10^2 + 11^2 + 12^2 + 13^2 + ... + 20^2) - (1^2 + 2^2 + 3^2 + ... + 10^2) - 10
= (20*21*41)/6 - (10*11*21)/6 - 10
= (10*7*41) - (5*11*7) -10
= 2475

Therefore, C is the answer.
Answer: C
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