Find a sequence $a$ so that $a_n = s \Delta a_n $.

Question

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

Any help with this question will be great. This is my first time doing recurrence relations and I am very confused.

Answer 1

With you definition, what is required is $$ a_n = sa_{n+1}-sa_n\iff a_{n+1} = \frac{s+1}s a_n $$ this is the definition of a geometric sequence.

The general solution has the form $$ a_n = A \left[ \frac{s+1}s \right]^n $$and with $1=a_0 = A$ you find that

$$ a_n = \left[ \frac{s+1}s \right]^n $$

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You have proved that IF $u$ was a solution it had the required form.

You then have to check that if you define $u$ this way, the relation holds. You see that you need the condition $s\neq 0$. Then $u$ is defined and:

$$ s\Delta a_n = sa_{n+1}-sa_n = sa_n \left[ \frac{s+1}s -1\right] = sa_n \frac{1}s = a_n $$