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

Question

If  (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}

ThatDudeKnows · Answer

(n+1)!*f(n) = (n-1)!  tells us that  f(n) = \frac{(n-1)!}{(n+1)!}

f(1)=\frac{(n-1)!}{(n+1)!}=\frac{0!}{2!}=\frac{1}{2}  To that, we are going to add more positive numbers, so we can eliminate answer choices A, B, and C.

The terms will be:
 \frac{(n-1)!}{(n+1)!}+\frac{(n)!}{(n+2)!}+\frac{(n+1)!}{(n+3)!}+\frac{(n+2)!}{(n+4)!}. . .

The denominator of the 100th term will be (n+100)!, so 101!. The 101 will need to stay in the denominator since it is prime and there is no other 101 that will appear in the numerator to offset it. Only one remaining answer choice has 101 in the denominator.

Answer choice E.

BrushMyQuant · Answer

(n+1)!*f(n) = (n-1)! and we need to find the value of f(1) + f(2) + … + f(100) ?

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

=> 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}{100} - \frac{1}{101}

Alternate terms will cancel out and we will be left with only the first and the last terms

=  1 - \frac{1}{101}  =  \frac{101 - 1}{101}  =  \frac{100}{101}

So,  Answer will be E 
Hope it helps!

Watch the following video to learn the Basics of Functions and Custom Characters

https://www.youtube.com/watch?v=b7JoMX0gk8M

Sonia2023 · Answer

chetan2u  - any other solution for this one?

chetan2u · Answer

Sonia2023 
Whenever I look at addition of different numbers and answer containing denominator of the last number f(100) here, I would think whether there is some possibility of all numbers getting cancelled except first and the last. 
This would generally happen when each term can be converted into 1/x - 1/(x+1).

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

Thus,
 f(1)=\frac{1}{1}-\frac{1}{2} 
 f(2)=\frac{1}{2}-\frac{1}{3} 
 f(3)=\frac{1}{3}-\frac{1}{4} 
.
.
 f(100)=\frac{1}{100}-\frac{1}{101}

When you add all of them, all terms except 1/1 and 1/101 are cancelled out. 
Sum =  1-\frac{1}{101}=\frac{100}{101}

Adit_ · Answer

The point to notice in the question is that value actually go till 1/100*101 and not till 1/99*100  . Always good to go through the question again once when ur done with the problem.

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

Prep Toolkit

Top 5 GMAT Debrief Videos

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards

GMAT Problem Solving (PS) Questions

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

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards