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16 Apr 2019, 01:30
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Difficulty:
55%
(hard)
Question Stats:
72% (03:19) correct
28%
(03:10)
wrong
based on 244
sessions
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[GMAT math practice question]
Given that \(1^2 + 2^2 + 3^2 + … + n^2 = \frac{n(n+1)(2n+1)}{6}\), what is the value of \(11^2(1-\frac{1}{11})(1+\frac{1}{11}) + 12^2(1-\frac{1}{12})(1+\frac{1}{12}) + 13^2(1-\frac{1}{13})(1+\frac{1}{13}) + … + 20^2(1-\frac{1}{20})(1+\frac{1}{20})?\)
\(A. 1080\)
\(B. 2405\)
\(C. 2475\)
\(D. 2880\)
\(E. 3600\)
Given that \(1^2 + 2^2 + 3^2 + … + n^2 = \frac{n(n+1)(2n+1)}{6}\), what is the value of \(11^2(1-\frac{1}{11})(1+\frac{1}{11}) + 12^2(1-\frac{1}{12})(1+\frac{1}{12}) + 13^2(1-\frac{1}{13})(1+\frac{1}{13}) + … + 20^2(1-\frac{1}{20})(1+\frac{1}{20})?\)
\(A. 1080\)
\(B. 2405\)
\(C. 2475\)
\(D. 2880\)
\(E. 3600\)
16 Apr 2019, 02:29
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\(11^2(1-\frac{1}{11})(1+\frac{1}{11}) = 11^2 (1 - \frac{1}{{11^2}}) = 11^2 - 1\)
\(12^2(1-\frac{1}{12})(1+\frac{1}{12}) = 12^2 - 1\)
... and so on.
We can rewrite this expression as: \(11^2 + 12^2 + ... + 20^2 -10\)
Since the sequence starts from 11, we can say that \(11^2 + 12^2 + ... + 20^2 = \frac{20(20+1)(40+1)}{6} - \frac{10(10 +1)(20+1)}{6}\).
Hence, solving this we get:
\(2870 - 385 - 10 = 2475\). Pick C.
\(12^2(1-\frac{1}{12})(1+\frac{1}{12}) = 12^2 - 1\)
... and so on.
We can rewrite this expression as: \(11^2 + 12^2 + ... + 20^2 -10\)
Since the sequence starts from 11, we can say that \(11^2 + 12^2 + ... + 20^2 = \frac{20(20+1)(40+1)}{6} - \frac{10(10 +1)(20+1)}{6}\).
Hence, solving this we get:
\(2870 - 385 - 10 = 2475\). Pick C.
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