# Find terms of a sequence and a formula

• Feb 16th 2010, 04:31 PM
summerset353
Find terms of a sequence and a formula
Let {a_n}=(1*3*5*...[(2n+1)/(2n!)]. Find the first five terms of this sequence and a simple formula for the ratio [a_(n+1)]/(a_n)]

I found the first five terms to be {a_1}=3/2, {a_2}=5/4, {a_3}=7/12, {a_4}=9/48, and {a_5}=11/140. Is this what I was supposed to do?
I know have to find a formula based on [a_(n+1)]/(a_n)]. How would I do that if a formula was already written, that being [(2n+1)/(2n!)].
• Feb 16th 2010, 04:36 PM
Drexel28
Quote:

Originally Posted by summerset353
Let {a_n}=(1*3*5*...[(2n+1)/(2n!)]. Find the first five terms of this sequence and a simple formula for the ratio [a_(n+1)]/(a_n)]

I found the first five terms to be {a_1}=3/2, {a_2}=5/4, {a_3}=7/12, {a_4}=9/48, and {a_5}=11/140. Is this what I was supposed to do?
I know have to find a formula based on [a_(n+1)]/(a_n)]. How would I do that if a formula was already written, that being [(2n+1)/(2n!)].

Clearly $\displaystyle \frac{a_{n+1}}{a_n}=\frac{1\cdot3\cdots(2n+1)\cdot (2n+3)}{(2n+2)!}\cdot\frac{(2n)!}{1\cdot3\cdots(2n +1)}=\frac{2n+3}{(2n+2)(2n+1)}$
• Feb 16th 2010, 04:52 PM
summerset353
What about the first five terms? Did I find them the right way?