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#### Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.  # Given that 1^2 + 2^2 + 3^2 + … + n^2 = n(n+1)(2n+1)/6, what is the

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Math Revolution GMAT Instructor V
Joined: 16 Aug 2015
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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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Difficulty:   65% (hard)

Question Stats: 63% (03:21) correct 38% (03:42) wrong based on 80 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 $$11^2(1-\frac{1}{11})(1+\frac{1}{11}) + 12^2(1-\frac{1}{12})(1+\frac{1}{12}) + 13^2(1-\frac{1}{13})(1+\frac{1}{13}) + … + 20^2(1-\frac{1}{20})(1+\frac{1}{20})?$$

$$A. 1080$$

$$B. 2405$$

$$C. 2475$$

$$D. 2880$$

$$E. 3600$$

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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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3
$$11^2(1-\frac{1}{11})(1+\frac{1}{11}) = 11^2 (1 - \frac{1}{{11^2}}) = 11^2 - 1$$

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

... 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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Joined: 18 Jan 2019
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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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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)
Manager  B
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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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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}
Manager  G
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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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1
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.
Math Revolution GMAT Instructor V
Joined: 16 Aug 2015
Posts: 9146
GMAT 1: 760 Q51 V42
GPA: 3.82
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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=>

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

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

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

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

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

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

$$= (10*7*41) – (5*11*7) -10$$

$$= 2475$$

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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  