# Sequence with factorial

• Mar 21st 2013, 10:21 PM
Newbie999
Sequence with factorial

Find showing all working, a recursive definition of the sequence with general term

$\displaystyle tn = 2(n+1)!5^n$ from n1 onwards

How am I suppose to find the recursive definition? What are the steps?
• Mar 21st 2013, 10:58 PM
Re: Sequence with factorial
A recursive definition requires two components
1) A $\displaystyle t_1$ (a base term)
2) A general formula relating $\displaystyle t_{n+1}$ and $\displaystyle t_{n}$ for $\displaystyle n \geq 1$ where n is a natural number
--

Finding $\displaystyle t_1$ is easy. Replace $\displaystyle n=1$ to get $\displaystyle t_1 = 2(1+1)!5^1 = 20$

Since we know the formula for $\displaystyle t_n$ we also know the formula for $\displaystyle t_{n+1}$ (replace n with n+1)
Now consider $\displaystyle \frac{t_{n+1}}{t_n}$, the ratio between two consecutive terms, and isolate $\displaystyle t_{n+1}$ for the recursive formula as desired. I'll leave the rest to you.
• Mar 21st 2013, 11:14 PM
Newbie999
Re: Sequence with factorial
Macsters thank you very much for the reply.

I don't think I came up with the right answer and I can't simply anymore because my algebra is very bad.

$\displaystyle tn+1/tn = \frac{2(n+2)!5^n}{2(n+1)!5^n}$

Simplified

$\displaystyle tn+1/tn = \frac{2n+24*5^n}{2n+2*5^n}$

I am a bit lost from this point. Can you please tell me if I am even on the right track?
• Mar 21st 2013, 11:25 PM
Re: Sequence with factorial
Quote:

Originally Posted by Newbie999
$\displaystyle tn+1/tn = \frac{2(n+2)!/5^n}{2(n+1)!5^n}$

is incorrect, but close. you also needed to replace n with n+1 in the exponent of 5 on the numerator

$\displaystyle \frac{t_{n+1}}{t_n} = \frac{2(n+2)!5^{n+1}}{2(n+1)!5^n}$

Notice that the 2's cancel out. and that
$\displaystyle \frac{(n+2)!}{(n+1)!} = \frac{(n+2)(n+1)!}{(n+1)!}} = (n+2)$ by definition of factorial
$\displaystyle \frac{5^{n+1}}{5^n} = \frac{5*5^n}{5^n} = 5$

hence

$\displaystyle \frac{t_{n+1}}{t_n} = \frac{2(n+2)!5^{n+1}}{2(n+1)!5^n} = 5(n+2)$
• Mar 21st 2013, 11:27 PM
Newbie999
Re: Sequence with factorial
Thank you sooo much : ) for everything.

Edit: One more question. I would like to send you a PM since this is solved but someone might found it useful to see the question here.

With the $\displaystyle tn+1/tn = \frac{5{^{n+1}}}{5^{n}}$
I pretty much understand everything above and the final answer.

But what if $\displaystyle tn-1/tn = \frac{5{^{n-1}}}{5^{n}}$

How can I calculate a sequence when it's relating to previous number? Lets say n-1 or n-2.
• Mar 22nd 2013, 09:00 AM
If the general formula for $\displaystyle t_n$ was $\displaystyle 5^n$ instead, then
$\displaystyle \frac{t_{n-1}}{t_n} = \frac{5^{n-1}}{5^n} = \frac{1}{5}$ hence $\displaystyle t_{n-1} = \frac{1}{5}t_n$
so this is still a relationship between two consecutive numbers. Replace n with n-1 and you get a relationship between $\displaystyle t_{n-1}$ and $\displaystyle t_{n-2}$