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Math Revolution GMAT Instructor V
Joined: 16 Aug 2015
Posts: 8731
GMAT 1: 760 Q51 V42
GPA: 3.82
(n+1)!f(n) = (n-1)!. What is the value of f(1) + f(2) + … + f(100)?  [#permalink]

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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$$f(1) + f(2) + … + f(100)$$?

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

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

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

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

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

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GMAT Club Legend  V
Joined: 18 Aug 2017
Posts: 6059
Location: India
Concentration: Sustainability, Marketing
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WE: Marketing (Energy and Utilities)
Re: (n+1)!f(n) = (n-1)!. What is the value of f(1) + f(2) + … + f(100)?  [#permalink]

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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
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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]

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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 $$= \frac{100}{101}$$

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Math Revolution GMAT Instructor V
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]

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=>

$$(n+1)!f(n) = (n-1)!$$

$$=> (n-1)!*n*(n+1)f(n) = (n-1)!$$

$$=> n*(n+1)f(n) = 1$$

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

$$=> f(n) = \frac{1}{n} – \frac{1}{(n+1)}$$

Thus $$f(1) + f(2) + … + f(100) = (\frac{1}{1} – \frac{1}{2}) + (\frac{1}{2} – \frac{1}{3}) + (\frac{1}{3}-\frac{1}{4}) + … + (\frac{1}{99} – \frac{1}{100}) + (\frac{1}{100} – \frac{1}{101}) = \frac{1}{1} – \frac{1}{101} = \frac{100}{101}$$

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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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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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