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

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

by Max@Math Revolution » Fri May 17, 2019 12:00 am

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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. 1/99
B. 1/100
C. 1/101
D. 99/100
E. 100/101
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by GMATGuruNY » Fri May 17, 2019 2:41 am
Max@Math Revolution wrote:[GMAT math practice question]

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

A. 1/99
B. 1/100
C. 1/101
D. 99/100
E. 100/101
f(n) = (n-1)!/(n+1)!

f(1) = 0!/2! = 1/2
f(2) = 1!/3! = 1/6
f(3) = 2!/4! = 1/12
f(4) = 3!/5! = 1/20

f(1) + f(2) = 1/2 + 1/6 = 2/3
f(1) + f(2) + f(3) = 2/3 + 1/12 = 3/4
f(1) + f(2) + f(3) + f(4) = 3/4 + 1/20 = 4/5

The sum of the first 2 terms = 2/3.
The sum of the first 3 terms = 3/4.
The sum of the first 4 terms = 4/5.
By extension:
The sum of the first 100 terms = 100/101.

The correct answer is E.
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by Max@Math Revolution » Sun May 19, 2019 4:57 pm
=>

(n+1)!f(n) = (n-1)!
=> (n-1)!*n*(n+1)f(n) = (n-1)!
=> n*(n+1)f(n) = 1
=> f(n) = 1/{n(n+1)}
=> f(n) = 1/n - 1/(n+1)

Thus f(1) + f(2) + ... + f(100) = (1/1 - 1/2) + (1/2 - 1/3) + (1/3-1/4) + ... + (1/99 - 1/100) + (1/100 - 1/101) = 1/1 - 1/101 = 100/101

Therefore, E is the answer.
Answer: E
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