Solving recurrence with non constant coefficients

Question

I am having a hard time to solve the following

$a_k=\left(\frac{d}{2}\right)^{k-2}a_{k-2}$ where $d$ is a parameter and $a_0=1$ $a_1=d$. Will appreciate your help.

Thanks!

Answer 1

To add to Did's answer.

$$ \Large{a_k = \prod_{n=1}^m(\frac{d}{2})^{k - 2n}a_{k - 2m} = (\frac{d}{2})^{\sum_{n=1}^m(k) - 2\sum_{n=1}^m (n)}a_{k - 2m}} $$

For the even case, notice that you can set $m = k/2$. So that the recursion ends at $a_0$.

The sums are then $$\sum_{n=1}^{k/2}(k) = k^2/2$$ and $$\sum_{n=1}^{k/2} (n) =\frac{k}{4}(\frac{k}{2} + 1).$$

So we have

$$\frac{k^2}{2} - 2\frac{k}{4}(\frac{k}{2} + 1) = \frac{k}{2}(\frac{k}{2} - 1) = \frac{k^2}{4} - \frac{k}{2}. $$

Thus for even $k$,

$$\Large{a_k = (\frac{d}{2})^{\frac{k^2}{4} - \frac{k}{2}}}.$$

Odd case is similar.

Answer 2

$$a_{2k}=a_0\cdot\left(\frac{d}2\right)^{k^2-k}\qquad a_{2k+1}=a_1\cdot\left(\frac{d}2\right)^{k^2}$$