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

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New post 17 May 2019, 00:10
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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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Re: (n+1)!f(n) = (n-1)!. What is the value of f(1) + f(2) + … + f(100)?  [#permalink]

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New post 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
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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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New post 17 May 2019, 02:50
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MathRevolution wrote:
[GMAT math practice question]

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

\(A. \frac{1}{99}\)

\(B. \frac{1}{100}\)

\(C. \frac{1}{101}\)

\(D. \frac{99}{100}\)

\(E. \frac{100}{101}\)


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

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

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

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

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New post 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
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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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New post 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?
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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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