((n-1)!+n!+(n+1)!)/n!

Question

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

Which of the following values of n will result between 16 and 17 for the equation?

A. 14
B. 15
C. 16
D. 17
E. 18

If I factor (n-1) and reduce the equation and then substitute the answer choices then result becomes a. But is there a shortcut to solving this.

KarishmaB · Accepted Answer

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

Note that out of the options, whatever value you give n, 1/n will be a very small term, much smaller than 1. 
In rest of the expression, you have 1 + (n + 1) which you want to be 16. So n must be 14.

Answer (A)

chetan2u · Answer

hi  jahanafsana ,
it depends how you have reduced the equation...
1) first is you open each term .... find common terms... get an equation and then substitute values..
2) second , which may be slightly less time consuming.....
 \frac{((n-1)!+n!+(n+1)!)}{n!} 
= \frac{(n-1)!}{n!} + \frac{n!}{n!} + \frac{(n+1)!}{n!} 
= \frac{1}{n} +1+n+1.... now you can easily see n+2 should be greater than or equal to 16.... 14 is the value...

CrackverbalGMAT · Answer

Easiest way is simplify the equation and then check the answer choice.
(n-1)! +n! + (n+1)! /n! = (I/n) +1+ (n+1)
(n+2) + (I/n) is 16 or more than that so only option is 14.

jahanafsana · Answer

Thank you for your reply. It helped.

jahanafsana · Answer

Thanks a lot for your detail reply. It is clear now.

saiesta · Answer

Could you show how you simplified this part in steps? I seem to get a different answer.

ENGRTOMBA2018 · Answer

You can adopt 2 methods to verify: either substitute n=3 and cross check. For algebraic method look below:

Realize that (n+1)! = (n+1)*n*(n-1)(n-2)... = (n+1)*n!

Similarly, n!=n*(n-1)! ---> (n-1)!= n!/n

Thus,  \frac{(n - 1)! + n! + (n + 1)!}{n!} =\frac{(n!/n)+n!+(n+1)n!}{n!}  = (cancelling n! from both the numerator and denominator you get)  \frac{1}{n}+1+(n+1)

Hope this helps.

chetan2u · Answer

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

KarishmaB · Answer

Note that 
 (n-1)! = 1*2*3*...*(n-1) 
 n! = 1*2*3*...*(n-1)*n

So n! just has an extra n in it.

When you divide (n-1)! by n!, you are just left with an n in the denominator.

Similarly, 
 n! = 1*2*3*...*(n-1)*n 
 (n+1)! = 1*2*3*...*(n-1)*n*(n+1) 
So (n+1)! only has an extra (n+1).

When you divide (n+1)! by n!, you are left with (n+1) in the denominator.

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

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

= \frac{1}{n} + 1 + (n+1)

shashankism · Answer

\frac{(n-1)!+n!+(n+1)!}{n!} 
=1/n+1+n+1
=n+2+(1/n)
16<n+2+1/n<17
->14<n+1/n<15
n=14
 Answer A.

BrentGMATPrepNow · Answer

Useful fraction property:  \frac{a+b+c}{d}=\frac{a}{d}+\frac{b}{d}+\frac{c}{d}

Apply the property to get:  \frac{(n-1)!+n!+(n+1)!}{n!}=\frac{(n-1)!}{n!}+\frac{n!}{n!}+\frac{(n+1)!}{n!}

=\frac{1}{n}+1+(n+1)

=n + 2 + \frac{1}{n}

From the answer choices, we can see that the value of n must be an integer from 14 to 18 inclusive, which means the fraction  \frac{1}{n}  must be between 0 and 1.

This means the expression  n + 2 + \frac{1}{n}  will be between 16 and 17 when  n = 14

Answer: A

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((n-1)!+n!+(n+1)!)/n!

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