Solve recurrence: a_n=s*a_{n+1}-s*a_{n}

**MSUMathStdnt** · March 15th 2011, 02:48 PM

I feel like I am missing something to this question.

Let $s$ be a real number, $s \ne 0$ . Find a sequence $a$ such that $a_n=s \Delta a_n$ and $a_0=1$ .

My answer (so far):
$a_n = s \Delta a_n = sa_{n+1}-sa_n\; \Rightarrow\; a_{n+1}=\frac{1+s}{s}a_n$ .

Changing indexes:

$a_n=\frac{1+s}{s}a_{n-1}$
$a_0=1$
$a_1=\left( \frac{1+s}{s} \right) \left( 1 \right) = \frac{1+s}{s}$
$a_2=\left( \frac{1+s}{s} \right) \left( \frac{1+s}{s} \right) = \left( \frac{1+s}{s} \right)^2$
$a_3=\left( \frac{1+s}{s} \right) \left( \frac{1+s}{s} \right)^2 = \left( \frac{1+s}{s} \right)^3$
$...$
$\mathbf{a_n= \left( \frac{1+s}{s} \right)^n}$

Is that really all they're looking for?
Thanks.

**topsquark** · March 15th 2011, 03:36 PM

Originally Posted by MSUMathStdnt

I feel like I am missing something to this question.

Let $s$ be a real number, $s \ne 0$ . Find a sequence $a$ such that $a_n=s \Delta a_n$ and $a_0=1$ .

My answer (so far):
$a_n = s \Delta a_n = sa_{n+1}-sa_n\; \Rightarrow\; a_{n+1}=\frac{1+s}{s}a_n$ .

Changing indexes:

$a_n=\frac{1+s}{s}a_{n-1}$
$a_0=1$
$a_1=\left( \frac{1+s}{s} \right) \left( 1 \right) = \frac{1+s}{s}$
$a_2=\left( \frac{1+s}{s} \right) \left( 1 \right) = \left( \frac{1+s}{s} \right)^2$
$a_3=\left( \frac{1+s}{s} \right) \left( 1 \right)^2 = \left( \frac{1+s}{s} \right)^3$
$...$
$\mathbf{a_n= \left( \frac{1+s}{s} \right)^n}$

Is that really all they're looking for?
Thanks.

Well, your notation for the a values is a little off, but your final formula is correct.

-Dan

**Soroban** · March 15th 2011, 03:39 PM

Hello, MSUMathStdnt!

$a_n \:=\:s\!\cdot\!a_{n+1} - s\!\cdot\!a_n,\;\;a_o = 1$

If there are no typos in the problem, the solution is elementary.

We have: . $a_n + s\!\cdot\!a_n \;=\;s\!\cdot\!a_{n+1} \quad\Rightarrow\quad (s+1)a_n \:=\:a_{n+1}$

. . Hence: . $a_{n+1} \:=\:\frac{s+1}{s}\,a_n$

Each term is a constant ratio of the preceding term.

. . We have a geometric series.

**MSUMathStdnt** · March 15th 2011, 08:53 PM

Originally Posted by topsquark

Well, your notation for the a values is a little off, but your final formula is correct.

-Dan

I think I fixed the notation you were referring to. I used copy and paste to type that up and forgot to change some lines.

Originally Posted by Soroban

Hello, MSUMathStdnt!

If there are no typos in the problem, the solution is elementary.

...

Each term is a constant ratio of the preceding term.

. . We have a geometric series.

Thanks, Soroban.

My book calls that a geometric sequence. The geometric series would be the sum, from $0$ to $n$ , of $a_n$ . The sum could then be replaced by:

$\frac{1-\left( \frac{1+s}{s} \right)^n}{1-\frac{1+s}{s}}=\left( \frac{1+s}{s} \right)^n-1$

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