Algebra with the Geometric Sequence formula

[Vertciel]

Hello there,

I am studying Sequences by myself and so far, I understand the concept. However, I do not understand how my book simplified the following equation:

\(\displaystyle a_n = 3(\frac{-2}{3}\)^{n-1}\)

\(\displaystyle = 3\frac{(-2)^{n-1}}{3^{n-1}}\)

\(\displaystyle = \frac{(-2)^{n-1}}{3^{n-2}}\)

Thank you in advance.

 

[Deleted member 4993]

Hello there,

I am studying Sequences by myself and so far, I understand the concept. However, I do not understand how my book simplified the following equation:

\(\displaystyle a_n = 3\frac{-2}{3}\^{n-1}\)

\(\displaystyle = 3\frac{(-2)^{n-1}}{3^{n-1}}\) ......\(\displaystyle because.... {(A/B)^{n}} = {(A)^{n}}/{B^{n}\)

\(\displaystyle = \frac{(-2)^{n-1}}{3^{n-2}}\) .....because.... \(\displaystyle \frac{(3)^{n-1}}{3} = {(3)^{n-2}}\)

Thank you in advance.

 

[Vertciel]

\(\displaystyle = \frac{(-2)^{n-1}}{3^{n-2}}\) .....because.... \(\displaystyle \frac{(3)^{n-1}}{3} = {(3)^{n-2}}\)

Thanks for your reply, Subhotosh Khan. Could you please explain in further detail the above equation?

 

[pka]

Do you understand the workings of exponents?
Those simply obey the very basic laws if exponents.

\(\displaystyle \L \frac{3}{{3^{n - 1} }} = \frac{1}{{3^{n - 2} }}\)

 

[Vertciel]

All right. Thanks for clearing that up.