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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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Math Revolution GMAT Instructor
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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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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"Only $79 for 1 month Online Course" "Free Resources-30 day online access & Diagnostic Test" "Unlimited Access to over 120 free video lessons - try it yourself" ##### Most Helpful Community Reply Manager Joined: 21 Feb 2019 Posts: 123 Location: Italy Re: Given that 1^2 + 2^2 + 3^2 + … + n^2 = n(n+1)(2n+1)/6, what is the [#permalink] ### Show Tags 16 Apr 2019, 01:29 2 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. ##### General Discussion Intern Joined: 18 Jan 2019 Posts: 28 Re: Given that 1^2 + 2^2 + 3^2 + … + n^2 = n(n+1)(2n+1)/6, what is the [#permalink] ### Show Tags 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) Manager Joined: 31 Dec 2018 Posts: 110 Location: India Re: Given that 1^2 + 2^2 + 3^2 + … + n^2 = n(n+1)(2n+1)/6, what is the [#permalink] ### Show Tags 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} Manager Joined: 21 Feb 2019 Posts: 123 Location: Italy Re: Given that 1^2 + 2^2 + 3^2 + … + n^2 = n(n+1)(2n+1)/6, what is the [#permalink] ### Show Tags 16 Apr 2019, 09:46 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 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] ### Show Tags 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 _________________ MathRevolution: Finish GMAT Quant Section with 10 minutes to spare The one-and-only World’s First Variable Approach for DS and IVY Approach for PS with ease, speed and accuracy. "Only$79 for 1 month Online Course"
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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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02 Jun 2020, 13:17
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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