# Find the smallest $n$ in series, such that $a_n>10^5$

I'm trying to solve the following problem:

We have a series $${a_n}^{\infty}_{n=0}$$ with relation: $$a_{n+2}=5a_{n+1}+3a_n$$; $$a_0=a_1=1$$. Find smallest $$n$$, such that $$a_n>10^5$$.

This is a preparation for exam - no calculators allowed. I tried to create a characteristic polynome, so I created an equation $$-x^2+5x+3=0$$, however I got roots $$x_1=\frac{5-\sqrt{37}}{2}, x_2=\frac{5+\sqrt{37}}{2}$$, which I think is wrong, as the numbers don't seem right for an exam.

Did I make a mistake somewhere when creating the equation or is there any other (simpler) way to solve this?

Thanks

I don't know what kind of test this is, but it's not that hard to math out approximately.

$$a_2=8$$, $$a_3=43$$, and from there let's assume that the function is $$a_{n+1}=5a_n$$. So $$a_4\approx200$$, $$a_5\approx1000$$, $$a_6\approx5000$$, $$a_7\approx25000$$, $$a_8\approx125000$$. Those are lower bounds, but certainly they're not so far off that $$a_7>10^5$$ could be true.

You can get an idea of the sequence by computing the first few terms exactly:

$$1,1,8,43,239$$

where $$a_4=239=5\cdot43+3\cdot8$$ is still easy to do by hand, or even mentally. At this point, some crude estimates can take over:

$$1000\lt a_5\lt1500$$

since $$1000=5\cdot200\lt5\cdot239+3\cdot43$$ and $$1500=5\cdot300\gt5\cdot239+3\cdot43$$, and then

$$5000\lt a_6\lt9000$$

since $$5000=5\cdot1000\lt5a_5\lt5a_5+3a_4$$ and $$9000=6\cdot1500\gt5\cdot1500+3\cdot239\gt5a_5+3a_4$$

From $$5a_6\lt5a_6+3a_5\lt8a_6$$, we now have

$$25000\lt a_7\lt72000$$

It now follows that $$a_8\gt5a_7\gt125000$$ is the first term greater than $$10^5$$.

Remark: the lower bounds here, $$1000, 5000, 25000, 125000$$, agree with those in Matthew Daly's answer (which posted while I was initially composing this answer). The upper bounds, $$1500,9000,72000$$, are the trickier ones to work out -- but only slightly trickier.