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

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 $a_0=1$.

$a_n = s \Delta a_n = sa_{n+1}-sa_n\; \Rightarrow\; a_{n+1}=\frac{1+s}{s}a_n$.

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.

2. 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 $a_0=1$.

$a_n = s \Delta a_n = sa_{n+1}-sa_n\; \Rightarrow\; a_{n+1}=\frac{1+s}{s}a_n$.

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

3. 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.

4. 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:

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