# Summation of a Series

• March 14th 2008, 12:18 PM
shelley808
Summation of a Series
I'm currently working on geometric sequences but I don't get something really simple about it.

My first question is:

How does (4^n)/[3^(n-1)] become 4[(4/3)^(n-1)]

and how does

(7^n)/[4^(2n-1)] become (4)[(7^n)/(4^2n)].

Thank you!!!
• March 14th 2008, 01:05 PM
Soroban
Hello, Shelley!

Quote:

How does . $\frac{4^n}{3^{n-1}}\:\text{ become }\:4\left(\frac{4}{3}\right)^{n-1}$

We have: . $\frac{4^n}{3^{n-1}} \;=\;\frac{4\cdot4^{n-1}}{3^{n-1}} \;=\;4\cdot\frac{4^{n-1}}{3^{n-1}} \;=\;4\left(\frac{4}{3}\right)^{n-1}$

Quote:

How does . $\frac{7^n}{4^{2n-1}}\:\text{ become }\:4\cdot\frac{7^n}{4^{2n}}$

We have: . $\frac{7^n}{4^{2n-1}} \;=\;\frac{7^n}{4^{2n}\cdot4^{-1}} \;=\;\frac{7^n}{4^{2n}\cdot\frac{1}{4}} \;=\;4\cdot\frac{7^n}{4^{2n}}$

Ha! . . . We can push it even further . . .

We have: . $4\cdot\frac{7^n}{(4^2)^n} \;=\;4\cdot\frac{7^n}{16^n} \;=\;7\left(\frac{7}{16}\right)^n
$

• March 14th 2008, 01:56 PM
shelley808
thank you so much!!!! however, i am still confused about one part, how does the 4^n become (4)(4^(n-1))?

thanks again!!
• March 14th 2008, 02:02 PM
Moo
Hello,

This is because $a^{b+c} = a^b a^c$

Hence as n=(n-1)+1, $4^n = 4^{(n-1)+1} = 4^1 4^{n-1} = 4 \times 4^{n-1}$