# Geometric Sequences and Series

• Jul 29th 2011, 04:45 PM
GrigOrig99
Geometric Sequences and Series
Q. Find Sn, the sum to n terms, of the geometric series 2 + 2/ 3 + 2/ 3^2 +…
If Sn = 242/ 81, find the value of n.

Eq. Sn = a (1 - r^n) / (1 - r) … (r < 1).

Attempt:
From the series above, r = (2/ 3) / 2 => 1/ 3
Therefore, a = 2 and r = 1/ 3. Sub these into the Sn equation…
2 (1 - (1/ 3)^n) / (1 - 1/3) =
2 (1 - (1/ 3)^n) / (2 / 3)

I’m stuck at this point, as the answer in the text book indicates that:

Sn = 3 - (1/ (3^(n-1)))

I am uncertain of the procedure they have applied to convert n into n - 1. Can anyone help? Thank you.
• Jul 29th 2011, 04:57 PM
pickslides
Re: Geometric Sequences and Series
Your working looks good, to finish it,

Solve for $n$ where $\displaystyle \frac{242}{81} = \frac{1-\frac{1}{3}^n}{\frac{2}{3}}$
• Jul 29th 2011, 05:09 PM
SammyS
Re: Geometric Sequences and Series
Quote:

Originally Posted by GrigOrig99
Q. Find Sn, the sum to n terms, of the geometric series 2 + 2/ 3 + 2/ 3^2 +…
If Sn = 242/ 81, find the value of n.

Eq. Sn = a (1 - r^n) / (1 - r) … (r < 1).
...

I’m stuck at this point, as the answer in the text book indicates that:

Sn = 3 - (1/ (3^(n-1)))

I am uncertain of the procedure they have applied to convert n into n - 1. Can anyone help? Thank you.

The 1st term in the series is (2)(1/3)^0 . It has an exponent of 0.

The 2nd is (2)(1/3)^1. It has an exponent of (2-1).
$\vdots$
The n-th term is 2)(1/3)^(n-1). It has an exponent of (n-1).
• Jul 29th 2011, 05:49 PM
GrigOrig99
Re: Geometric Sequences and Series
Ok, so following on from:
2 (1 - (1/ 3)^n) / (2 / 3)

I have done the following:

= (2(1) - 2(1/ 3)^n) / (2/ 3)
= (2 - (2/3)^n) / (2/ 3)
= (2(3/ 2) - (3/ 2)((2/ 3)^n))
= (6/ 2 - (6/ 6)^n)
= 3 - 1^(n-1)

Recall that the text book answer is Sn = 3 - (1/ (3^(n-1))).

I guess I'm closer to the mark now, but I still don't have it entirely correct. Can anyone see where I've gone wrong? Thanks again.
• Jul 29th 2011, 06:23 PM
SammyS
Re: Geometric Sequences and Series
Quote:

Originally Posted by GrigOrig99
Ok, so following on from:
2 (1 - (1/ 3)^n) / (2 / 3)

I have done the following:

= (2(1) - 2(1/ 3)^n) / (2/ 3)
You're good to the above line.

= (2 - (2/3)^n) / (2/ 3)
(2)(1/3)^n ≠ (2/3)^n)

= (2(3/ 2) - (3/ 2)((2/ 3)^n))
= (6/ 2 - (6/ 6)^n)
= 3 - 1^(n-1)

Recall that the text book answer is Sn = 3 - (1/ (3^(n-1))).

I guess I'm closer to the mark now, but I still don't have it entirely correct. Can anyone see where I've gone wrong? Thanks again.

$2\ \left(\frac{1 - (1/ 3)^n}{2 / 3}\right)= \frac{2\cdot 3(1 - (1/ 3)^n)}{2}\right=3-(1/3)^{(n-1)}$

Notice that $\frac{242}{81}=\frac{243}{81}-\frac{1}{81}=3-\frac{1}{81}$
• Jul 29th 2011, 07:13 PM
GrigOrig99
Re: Geometric Sequences and Series
Ok, I'm almost clear on this. So, is it correct to say that for:

(2.3 (1 - (1/3)^n)/ 2

we do 'not' actually multiply a number (1/3^n, in this case) by another (i.e., 2.3 or 6) when it is expressed to the power of an unknown (n)?
• Jul 30th 2011, 12:58 AM
Siron
Re: Geometric Sequences and Series
Yes, but we can mulitply them if the power's base is the same as the constant. For example: $3\cdot 3^x=3^{x+1}$. (I don't know if this answers your question).

Have you found the value of n now? ...
• Jul 30th 2011, 03:24 AM
GrigOrig99
Re: Geometric Sequences and Series
Great, thanks for clearing that up. n should be 5, then.

Thanks again to everyone who helped.
• Jul 30th 2011, 07:39 AM
Siron
Re: Geometric Sequences and Series
Right! n=5