Solve Geometric Series: Find n from s=9-32-n

[ChelseaL]

Given that the sum of the first n terms of series, s, is 9-3_2-n
show that the s is a geometric progression.

Do I use the formula a_n = a_r ^n-1? And if so, how do I apply it?

 

[ChelseaL]

So far I have this

9-3^2-n

3 = 9
= 9 ^-(n-2)
= 9 ^-(n-1-1)
= 9 ^{-(n-1) + 1}
= 9 \cdot 3 ^-(n-1)
= 9 (\frac{1}{3}) n-1
where a=9 and r=\frac{1}{3}

 

[MarkFL]

We have:

\(\displaystyle S_{n}=S_{n-1}+a_{n}\implies a_{n}=S_{n}-S_{n-1}=\left(9-3^{2-n}\right)-\left(9-3^{2-(n-1)}\right)=\frac{18}{3^n}\)

To show that $a_n$ is a GP, we need to show that the ratio \(\displaystyle \frac{a_{n+1}}{a_{n}}\) is a constant, that is, not dependent on $n$.

Can you proceed?

 

[ChelseaL]

What do you mean?

 

[skeeter]

What do you mean?

show the ratio $\dfrac{a_{n+1}}{a_n}$ is a constant

 

[ChelseaL]

18/3_n + 1 / (18/3ⁿ)?

Sorry for not really understanding this topic. My lecturer regularly misses class and refuses to help us when we ask questions.

 

[MarkFL]

18/3_n + 1 / (18/3ⁿ)?

Sorry for not really understanding this topic. My lecturer regularly misses class and refuses to help us when we ask questions.

If \(\displaystyle a_n=\frac{18}{3^n}\) then \(\displaystyle a_{n+1}=\frac{18}{3^{n+1}}\) and so:

\(\displaystyle \frac{a_{n+1}}{a_{n}}=\frac{\dfrac{18}{3^{n+1}}}{\dfrac{18}{3^{n}}}=\,?\)

 

[ChelseaL]

\frac{1}{n+1}?

 

[MarkFL]

\frac{1}{n+1}?

No, I think you are not using proper rules of algebra. Observe that:

\(\displaystyle a_{n+1}=\frac{18}{3^{n+1}}=\frac{18}{3\cdot3^n}=\frac{1}{3}a_n\)

Hence:

\(\displaystyle \frac{a_{n+1}}{a_{n}}=\frac{1}{3}\)

 

[ChelseaL]

ohhhh. What do I need to do from here?

 

[MarkFL]

ohhhh. What do I need to do from here?

As far as this problem goes, it is done, because we have shown the ratio is a constant. :)