# Math Help - Sequence with factorial

1. ## Sequence with factorial

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

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

How am I suppose to find the recursive definition? What are the steps?

2. ## Re: Sequence with factorial

A recursive definition requires two components
1) A $t_1$ (a base term)
2) A general formula relating $t_{n+1}$ and $t_{n}$ for $n \geq 1$ where n is a natural number
--

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

Since we know the formula for $t_n$ we also know the formula for $t_{n+1}$ (replace n with n+1)
Now consider $\frac{t_{n+1}}{t_n}$, the ratio between two consecutive terms, and isolate $t_{n+1}$ for the recursive formula as desired. I'll leave the rest to you.

3. ## 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.

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

Simplified

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

I am a bit lost from this point. Can you please tell me if I am even on the right track?

4. ## Re: Sequence with factorial

Originally Posted by Newbie999
$tn+1/tn = \frac{2(n+2)!/5^n}{2(n+1)!5^n}$
is incorrect, but close. you also needed to replace n with n+1 in the exponent of 5 on the numerator

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

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

hence

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

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

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

How can I calculate a sequence when it's relating to previous number? Lets say n-1 or n-2.

6. ## Re: Sequence with factorial

If the general formula for $t_n$ was $5^n$ instead, then

$\frac{t_{n-1}}{t_n} = \frac{5^{n-1}}{5^n} = \frac{1}{5}$ hence $t_{n-1} = \frac{1}{5}t_n$
so this is still a relationship between two consecutive numbers. Replace n with n-1 and you get a relationship between $t_{n-1}$ and $t_{n-2}$