Solve the following recurrence $b(n)=0.9*b(n-1)+60, b(1)=300$

Question

How to solve the following recurrence: $$b(n)=0.9*b(n-1)+60, b(1)=300$$

I do it with telescoping, like this:

$$b(n)=0.9*(0.9*b(n-2)+60)+60$$ $$b(n)=0.9*(0.9*(0.9*b(n-3)+60)+60)+60$$

Which means that for $n=4$ we get $$b(n)=0.9^{(n-1)}*300+(60)^n-1$$

But the correct solution is $b(n)=-300*(0.9^n)+600$

Can someone please explain where is my mistake?

Answer 1

Let $b_n = ab_{n-1} + c, b_1 = d$. Then by expanding, one gets: $b_n = a(ab_{n-2} + c) + c = a^{n-1}b_1 + \sum_{i=0}^{n-1}{a^ic}$. Finally $b_n =a^{n-1}d + c\frac{a^n-1}{a-1}$.

Answer 2

Let $$b_{n}=a_n+c$$ So we have that $$a_n+c=0.9(a_{n-1}+c)+60$$ $$a_n=0.9a_{n-1}+60-0.1c$$ So if we let $60-0.1c=0$, we get that $c=600$. But this means that $$a_n=0.9a_{n-1}$$ Which I think you can solve. From this, we will get that $$b_n=a_n+600$$