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# (n+1)!f(n) = (n-1)!. What is the value of f(1) + f(2) + … + f(100)?

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Math Revolution GMAT Instructor
Joined: 16 Aug 2015
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(n+1)!f(n) = (n-1)!. What is the value of f(1) + f(2) + … + f(100)?  [#permalink]

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17 May 2019, 00:10
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Difficulty:

65% (hard)

Question Stats:

51% (01:51) correct 49% (02:30) wrong based on 55 sessions

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[GMAT math practice question]

$$(n+1)!f(n) = (n-1)!.$$ What is the value of$$f(1) + f(2) + … + f(100)$$?

$$A. \frac{1}{99}$$

$$B. \frac{1}{100}$$

$$C. \frac{1}{101}$$

$$D. \frac{99}{100}$$

$$E. \frac{100}{101}$$

_________________
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" "Free Resources-30 day online access & Diagnostic Test" "Unlimited Access to over 120 free video lessons - try it yourself" GMAT Club Legend Joined: 18 Aug 2017 Posts: 6059 Location: India Concentration: Sustainability, Marketing GPA: 4 WE: Marketing (Energy and Utilities) Re: (n+1)!f(n) = (n-1)!. What is the value of f(1) + f(2) + … + f(100)? [#permalink] ### Show Tags 17 May 2019, 00:29 MathRevolution wrote: [GMAT math practice question] $$(n+1)!f(n) = (n-1)!.$$ What is the value of$$f(1) + f(2) + … + f(100)$$? $$A. \frac{1}{99}$$ $$B. \frac{1}{100}$$ $$C. \frac{1}{101}$$ $$D. \frac{99}{100}$$ $$E. \frac{100}{101}$$ $$(n+1)!f(n) = (n-1)!.$$ (n+1)*n*(n-1)! f(n)= (n-1)! or say f(n)=1/(n+1)*n f1=1/2 f2=1/6 f3=1/12 f4=1/20 f(5)=1/30 . . . .f(100)= 1/100*101 so we can write; 1/2+1/6+1/12+1/20+...+1/100*101 1/2*(1+1/3+1/6+1/20+...+1/50*101) solving the eqn will give us dr with 101 and nr sum ~100 ; since total terms are 100 IMOE Director Joined: 04 Aug 2010 Posts: 547 Schools: Dartmouth College Re: (n+1)!f(n) = (n-1)!. What is the value of f(1) + f(2) + … + f(100)? [#permalink] ### Show Tags 17 May 2019, 02:50 3 MathRevolution wrote: [GMAT math practice question] $$(n+1)!f(n) = (n-1)!.$$ What is the value of$$f(1) + f(2) + … + f(100)$$? $$A. \frac{1}{99}$$ $$B. \frac{1}{100}$$ $$C. \frac{1}{101}$$ $$D. \frac{99}{100}$$ $$E. \frac{100}{101}$$ $$f(n) = \frac{(n-1)!}{(n+1)!}$$ $$f(1) = \frac{0!}{2!} = \frac{1}{2}$$ $$f(2) = \frac{1!}{3!} = \frac{1}{6}$$ $$f(3) = \frac{2!}{4!} = \frac{1}{12}$$ $$f(4) = \frac{3!}{5!} = \frac{1}{20}$$ $$f(1) + f(2) = \frac{1}{2} + \frac{1}{6} = \frac{2}{3}$$ $$f(1) + f(2) + f(3) = \frac{2}{3} + \frac{1}{12} = \frac{3}{4}$$ $$f(1) + f(2) + f(3) + f(4) = \frac{3}{4} + \frac{1}{20} = \frac{4}{5}$$ The sum of the first 2 terms $$= \frac{2}{3}$$ The sum of the first 3 terms $$= \frac{3}{4}$$ The sum of the first 4 terms $$= \frac{4}{5}$$ By extension: The sum of the first 100 terms $$= \frac{100}{101}$$ _________________ GMAT and GRE Tutor New York, NY Available for tutoring in NYC and long-distance. For more information, please email me at GMATGuruNY@gmail.com. Math Revolution GMAT Instructor Joined: 16 Aug 2015 Posts: 8731 GMAT 1: 760 Q51 V42 GPA: 3.82 Re: (n+1)!f(n) = (n-1)!. What is the value of f(1) + f(2) + … + f(100)? [#permalink] ### Show Tags 19 May 2019, 17:07 => $$(n+1)!f(n) = (n-1)!$$ $$=> (n-1)!*n*(n+1)f(n) = (n-1)!$$ $$=> n*(n+1)f(n) = 1$$ $$=> f(n) = \frac{1}{{n(n+1)}}$$ $$=> f(n) = \frac{1}{n} – \frac{1}{(n+1)}$$ Thus $$f(1) + f(2) + … + f(100) = (\frac{1}{1} – \frac{1}{2}) + (\frac{1}{2} – \frac{1}{3}) + (\frac{1}{3}-\frac{1}{4}) + … + (\frac{1}{99} – \frac{1}{100}) + (\frac{1}{100} – \frac{1}{101}) = \frac{1}{1} – \frac{1}{101} = \frac{100}{101}$$ Therefore, E is the answer. Answer: E _________________ 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: (n+1)!f(n) = (n-1)!. What is the value of f(1) + f(2) + … + f(100)?  [#permalink]

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10 Jul 2019, 09:49
GMATGuruNY wrote:
MathRevolution wrote:
[GMAT math practice question]

$$(n+1)!f(n) = (n-1)!.$$ What is the value of$$f(1) + f(2) + … + f(100)$$?

$$A. \frac{1}{99}$$

$$B. \frac{1}{100}$$

$$C. \frac{1}{101}$$

$$D. \frac{99}{100}$$

$$E. \frac{100}{101}$$

$$f(n) = \frac{(n-1)!}{(n+1)!}$$

$$f(1) = \frac{0!}{2!} = \frac{1}{2}$$
$$f(2) = \frac{1!}{3!} = \frac{1}{6}$$
$$f(3) = \frac{2!}{4!} = \frac{1}{12}$$
$$f(4) = \frac{3!}{5!} = \frac{1}{20}$$

$$f(1) + f(2) = \frac{1}{2} + \frac{1}{6} = \frac{2}{3}$$
$$f(1) + f(2) + f(3) = \frac{2}{3} + \frac{1}{12} = \frac{3}{4}$$
$$f(1) + f(2) + f(3) + f(4) = \frac{3}{4} + \frac{1}{20} = \frac{4}{5}$$

The sum of the first 2 terms $$= \frac{2}{3}$$
The sum of the first 3 terms $$= \frac{3}{4}$$
The sum of the first 4 terms $$= \frac{4}{5}$$
By extension:
The sum of the first 100 terms $$= \frac{100}{101}$$

Hi, can you suggest me an approach to tackling such questions, like any tip i can implement to identify such hidden patterns?
Re: (n+1)!f(n) = (n-1)!. What is the value of f(1) + f(2) + … + f(100)?   [#permalink] 10 Jul 2019, 09:49
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