College Algebra 7th Edition

Published by Brooks Cole
ISBN 10: 1305115546
ISBN 13: 978-1-30511-554-5

Chapter 8, Sequences and Series - Chapter 8 Review - Exercises: 5

Answer

$\displaystyle a_1=1$ $\displaystyle a_2=3$ $\displaystyle a_3=15$ $\displaystyle a_4=105$ $\displaystyle a_{10}=654729075$

Work Step by Step

We are given: $a_{n}= \frac{(2n)!}{2^{n}n!}$ We evaluate: $\displaystyle a_1 = \frac{(2*1)!}{2^{1}1!}=\frac{2}{2}=1$ $\displaystyle a_2= \frac{(2*2)!}{2^{2}2!}=\frac{4*3*2!}{4*2!}=3$ $\displaystyle a_3= \frac{(2*3)!}{2^{3}3!}=\frac{6*5*4*3!}{8*3!}=15$ $\displaystyle a_4= \frac{(2*4)!}{2^{4}4!}=\frac{8*7*6*5*4!}{16*4!}=105$ $\displaystyle a_{10}= \frac{(2*10)!}{2^{10}*10!}=\frac{20!}{2^{10}*10!}=654729075$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.