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