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

AI Thread Summary
The discussion focuses on demonstrating that the series defined by the sum s = 9 - 32 - n is a geometric progression. Participants explore the application of the formula for the nth term of a geometric series and derive the terms a_n and a_{n+1}. They confirm that the ratio a_{n+1}/a_n is constant, specifically 1/3, which is a key characteristic of geometric progressions. The conversation highlights confusion around algebraic manipulation, but ultimately concludes that the problem is resolved by establishing the constant ratio. The final agreement indicates that the proof is complete.
ChelseaL
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Given that the sum of the first n terms of series, s, is 9-32-n
show that the s is a geometric progression.

Do I use the formula an = ar n-1? And if so, how do I apply it?
 
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So far I have this

9-32-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}
 
We have:

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

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

Can you proceed?
 
What do you mean?
 
ChelseaL said:
What do you mean?

show the ratio $\dfrac{a_{n+1}}{a_n}$ is a constant
 
18/3n + 1 / (18/3n)?

Sorry for not really understanding this topic. My lecturer regularly misses class and refuses to help us when we ask questions.
 
ChelseaL said:
18/3n + 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 $$a_n=\frac{18}{3^n}$$ then $$a_{n+1}=\frac{18}{3^{n+1}}$$ and so:

$$\frac{a_{n+1}}{a_{n}}=\frac{\dfrac{18}{3^{n+1}}}{\dfrac{18}{3^{n}}}=\,?$$
 
\frac{1}{n+1}?
 
ChelseaL said:
\frac{1}{n+1}?

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

$$a_{n+1}=\frac{18}{3^{n+1}}=\frac{18}{3\cdot3^n}=\frac{1}{3}a_n$$

Hence:

$$\frac{a_{n+1}}{a_{n}}=\frac{1}{3}$$
 
  • #10
ohhhh. What do I need to do from here?
 
  • #11
ChelseaL said:
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. :)
 

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