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!!!

Printable View

- Mar 14th 2008, 11:18 AMshelley808Summation 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!!! - Mar 14th 2008, 12:05 PMSoroban
Hello, Shelley!

Quote:

How does .

We have: .

Quote:

How does .

We have: .

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

We have: .

- Mar 14th 2008, 12:56 PMshelley808
thank you so much!!!! however, i am still confused about one part, how does the 4^n become (4)(4^(n-1))?

thanks again!! - Mar 14th 2008, 01:02 PMMoo
Hello,

This is because

Hence as n=(n-1)+1,