• Oct 13th 2008, 09:47 PM
consider the following geometric sequence 400, 320, 256, 204.8, ......

i) what is the recurrance system that descibes this sequence?
( denote the sequence by xn, and its first term by x1.)

ii) find the closed form for this sequence.

iii) use the closed form from (ii) to find the tenth term of the sequence, giving your answer correct to 4 d.p.
• Oct 14th 2008, 12:36 AM
Prove It
Quote:

consider the following geometric sequence 400, 320, 256, 204.8, ......

i) what is the recurrance system that descibes this sequence?
( denote the sequence by xn, and its first term by x1.)

ii) find the closed form for this sequence.

iii) use the closed form from (ii) to find the tenth term of the sequence, giving your answer correct to 4 d.p.

The common ratio is $\frac{400}{320} = \frac{5}{4}$.

Each term therefore is $x_{n+1} = x_nr = \frac{5}{4}x_n$.

So $x_2 = \frac{5}{4}x_1$

$x_3 = \frac{5}{4}x_2 = \frac{5}{4}\frac{5}{4}x_2 = \left(\frac{5}{4}\right)^2x_1$

$x_4 = \frac{5}{4}x_3 = \frac{5}{4}\left(\frac{5}{4}\right)^2 x_2 = \left(\frac{5}{4}\right)^3x_1$.

So in terms of $x_1$,

$x_{n+1}=\left(\frac{5}{4}\right)^n x_n$.
• Oct 14th 2008, 12:48 AM
Prove It
Quote:

consider the following geometric sequence 400, 320, 256, 204.8, ......

i) what is the recurrance system that descibes this sequence?
( denote the sequence by xn, and its first term by x1.)

ii) find the closed form for this sequence.

iii) use the closed form from (ii) to find the tenth term of the sequence, giving your answer correct to 4 d.p.

Supposing that we had a series

$S_n = x_1 + x_1 r + x_1 r^2 + \dots x_1 r^n$.

Multiply this whole series by r, we get

$rS_n = x_1 r + x_1 r^2 + x_1 r^3 + \dots x_1 r^{n+1}$.

Subtract $S_n$ from $rS_n$ and we get

$rS_n - S_n = x_1 r^{n+1} - x_1$

$S_n(r - 1) = x_1 (r^{n+1} - 1)$

$S_n = \frac{x_1 (r^{n+1} - 1)}{r - 1}$.

So if $x_1 = 400$ and $r = \frac{5}{4}$ the closed form for this series is

$S_n = \frac{400 \left(\frac{5}{4}^{n+1} - 1\right)}{\frac{5}{4} - 1}$.

$S_n = \frac{400 \left(\frac{5}{4}^{n+1} - 1\right)}{\frac{1}{4}}$

$S_n = 1600 \left(\frac{5}{4}^{n+1} - 1\right)$.
• Feb 22nd 2009, 02:56 PM
NitrousUK
Tisk tisk Toplad, shouldn't be posting your Open University Mathematics coursework questions on forums ;)
• Feb 22nd 2009, 04:08 PM
mr fantastic
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