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Given that 1^2 + 2^2 + 3^2 + … + n^2 = n(n+1)(2n+1)/6, what is the

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Given that 1^2 + 2^2 + 3^2 + … + n^2 = n(n+1)(2n+1)/6, what is the  [#permalink]

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New post 16 Apr 2019, 00:30
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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 \(11^2(1-\frac{1}{11})(1+\frac{1}{11}) + 12^2(1-\frac{1}{12})(1+\frac{1}{12}) + 13^2(1-\frac{1}{13})(1+\frac{1}{13}) + … + 20^2(1-\frac{1}{20})(1+\frac{1}{20})?\)

\(A. 1080\)

\(B. 2405\)

\(C. 2475\)

\(D. 2880\)

\(E. 3600\)

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Re: Given that 1^2 + 2^2 + 3^2 + … + n^2 = n(n+1)(2n+1)/6, what is the  [#permalink]

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New post 16 Apr 2019, 01:29
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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.
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Re: Given that 1^2 + 2^2 + 3^2 + … + n^2 = n(n+1)(2n+1)/6, what is the  [#permalink]

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New post 16 Apr 2019, 04:54
lucajava wrote:
\(11^2(1-\frac{1}{11})(1+\frac{1}{11}) = 11^2 (1 - \frac{1}{{11^2}}) = 11^2 - 1\)

\(12^2(1-\frac{1}{12})(1+\frac{1}{12}) = 12^2 - 1\)

... 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.


why is it not: (20(20+1)(40+1)6)−(11(11+1)(22+1)/6)
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Re: Given that 1^2 + 2^2 + 3^2 + … + n^2 = n(n+1)(2n+1)/6, what is the  [#permalink]

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New post 16 Apr 2019, 05:25
lucajava wrote:
\(11^2(1-\frac{1}{11})(1+\frac{1}{11}) = 11^2 (1 - \frac{1}{{11^2}}) = 11^2 - 1\)

\(12^2(1-\frac{1}{12})(1+\frac{1}{12}) = 12^2 - 1\)

... 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.



Please explain how you got the other half : \frac{10(10 +1)(20+1)}{6}
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Re: Given that 1^2 + 2^2 + 3^2 + … + n^2 = n(n+1)(2n+1)/6, what is the  [#permalink]

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New post 16 Apr 2019, 09:46
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Ish1996 because you are subtracting the sequence which starts from 1 to 10. So, \(n = 10\).

mohitranjan05: from the question, you are interested about the value of sequence \(11^2 + 12^2 + ... + 20^2\). Applying the given formula for \(n = 20\), you need to subtract the sequence \(1^2 + 2^2 + ... + 10^2\) to get what you're finding out.
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Re: Given that 1^2 + 2^2 + 3^2 + … + n^2 = n(n+1)(2n+1)/6, what is the  [#permalink]

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New post 18 Apr 2019, 00:05
=>

\(11^2(1-\frac{1}{11})(1+\frac{1}{11}) + 12^2(1-\frac{1}{12})(1+\frac{1}{12}) + 13^2(1-\frac{1}{13})(1+\frac{1}{13}) + … + 20^2(1-\frac{1}{20})(1+\frac{1}{20})\)

\(= 11^2(1-\frac{1}{11^2}) + 12^2(1-\frac{1}{12^2}) + 13^2(1-\frac{1}{13^2}) + … + 20^2(1-\frac{1}{20^2})\)

\(= (11^2-1) + (12^2-1) + (13^2-1) + … + (20^2-1)\)

\(= (11^2 + 12^2 + 13^2 + … + 20^2) - 10\)

\(= (1^2 + 2^2 + 3^2 + … + 10^2 + 11^2 + 12^2 + 13^2 + … + 20^2) - (1^2 + 2^2 + 3^2 + … + 10^2) – 10\)

\(= \frac{(20*21*41)}{6} – \frac{(10*11*21)}{6} – 10\)

\(= (10*7*41) – (5*11*7) -10\)

\(= 2475\)

Therefore, C is the answer.
Answer: C
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Re: Given that 1^2 + 2^2 + 3^2 + … + n^2 = n(n+1)(2n+1)/6, what is the  [#permalink]

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Re: Given that 1^2 + 2^2 + 3^2 + … + n^2 = n(n+1)(2n+1)/6, what is the   [#permalink] 02 Jun 2020, 13:17

Given that 1^2 + 2^2 + 3^2 + … + n^2 = n(n+1)(2n+1)/6, what is the

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