saiesta wrote:

VeritasPrepKarishma wrote:

jahanafsana wrote:

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

n!

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

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.

a. 14

b. 15

c. 16

d. 17

e. 18

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

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

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

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

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

_________________

Karishma

Veritas Prep GMAT Instructor

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