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100 points for $49 worth of Veritas practice GMATs FREE VERITAS PRACTICE GMAT EXAMS Earn 10 Points Per Post Earn 10 Points Per Thanks Earn 10 Points Per Upvote ## For all positive integers n, the sequence A_n is defined by tagged by: AAPL ##### This topic has 1 expert reply and 1 member reply ### Top Member ## For all positive integers n, the sequence A_n is defined by ## Timer 00:00 ## Your Answer A B C D E ## Global Stats Difficult Manhattan Prep $$\text{For all positive integers n, the sequence } A_n \text{ is defined by the following relationship:}$$ $$A_n=\frac{n-1}{n!}$$ $$\text{What is the sum of all the terms in the sequence from } A_1 \text{ through } A_{10}, \text{ inclusive?}$$ $$\text{A. }\frac{9!+1}{10!}$$ $$\text{B. }\frac{9(9!)}{10!}$$ $$\text{C. }\frac{10!-1}{10!}$$ $$\text{D. }\frac{10!}{10!+1}$$ $$\text{E. }\frac{10(10!)}{11!}$$ OA C ### Top Member Master | Next Rank: 500 Posts Joined 15 Oct 2009 Posted: 318 messages Upvotes: 27 AAPL wrote: Manhattan Prep $$\text{For all positive integers n, the sequence } A_n \text{ is defined by the following relationship:}$$ $$A_n=\frac{n-1}{n!}$$ $$\text{What is the sum of all the terms in the sequence from } A_1 \text{ through } A_{10}, \text{ inclusive?}$$ $$\text{A. }\frac{9!+1}{10!}$$ $$\text{B. }\frac{9(9!)}{10!}$$ $$\text{C. }\frac{10!-1}{10!}$$ $$\text{D. }\frac{10!}{10!+1}$$ $$\text{E. }\frac{10(10!)}{11!}$$ OA C Try developing a pattern. First number in sequence is (1-1)/1 = 0 second is 1/2 third is (3-1)/3! = 1/3 The sum of this series is 1/2 + 1/3 = 5/6 Test some answers recognizing that since 3 numbers in the sequence, 2 corresponds to 9, 3 corresponds to 10 and 4 to 11 Try A: ( 2!+ 1)/3! = 3/6 =1/2 > no match B: 2(2!)/3! = 4/6=2/3 > no match C: (3!-1)/3! = (6-1)/6 = 5/6 > We have a winner ### GMAT/MBA Expert GMAT Instructor Joined 25 Apr 2015 Posted: 1651 messages Followed by: 14 members Upvotes: 43 AAPL wrote: Manhattan Prep $$\text{For all positive integers n, the sequence } A_n \text{ is defined by the following relationship:}$$ $$A_n=\frac{n-1}{n!}$$ $$\text{What is the sum of all the terms in the sequence from } A_1 \text{ through } A_{10}, \text{ inclusive?}$$ $$\text{A. }\frac{9!+1}{10!}$$ $$\text{B. }\frac{9(9!)}{10!}$$ $$\text{C. }\frac{10!-1}{10!}$$ $$\text{D. }\frac{10!}{10!+1}$$ $$\text{E. }\frac{10(10!)}{11!}$$ OA C Notice that (n - 1)/n! = n/n! - 1/n! = 1/(n - 1)! - 1/n!. Thus: A1 + A2 + A3 + â€¦ + A10 = 1/0! - 1/1! + 1/1! - 1/2! + 1/2! - 1/3! + â€¦ + 1/9! - 1/10!. All the terms in the middle will cancel with either the preceding or the succeeding term; therefore we are left with: 1/0! - 1/10! = 1 - 1/10! = (10! - 1)/10! Answer: C _________________ Scott Woodbury-Stewart Founder and CEO • Free Trial & Practice Exam BEAT THE GMAT EXCLUSIVE Available with Beat the GMAT members only code • Free Practice Test & Review How would you score if you took the GMAT Available with Beat the GMAT members only code • FREE GMAT Exam Know how you'd score today for$0

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