Proof by induction of a regression formula (series)

Question

As part of the curriculum, we have recently been working on proofs by induction. During the exercises I encountered an exercise that I do not know how to solve, I would be happy to help. Sorry beforehand that I am attaching a picture, I could not raise the question as a built-in code on the site.

problem To the best of my understanding, there is a need to prove by induction of n, on the one hand, and on the other hand to prove that there is n0 so that the argument will be correct. I'm pretty lost..

I tried to prove for N + 1 organs, but then I did not understand how to proceed to exclude the existence of the n0 requested .

Any help will be appreciated

Answer 1

I guess you could write: $a_{n+1}=2a_n + a_{n-1}$. Then you apply the induction hypothesis $a_n \leq 2a_{n-1} + a_{n-2}$. This gives $$a_{n+1} \leq 2[(\frac{5}{2})^n +(\frac{5}{2})^{n-1}] = \frac{5^{n} + 2\cdot 5^{n-1}}{2^{n+1}} \leq (\frac{5}{2})^{n+1}$$ Using the $n_0$ Noble Mushtak found, this should complete the proof by induction.

Answer 2

You can find the value of $n_0$ by guess and check.

$$n=1\implies a_n=6 > \left(\frac 5 2\right)^n=2.5$$ $$n=2\implies a_n=7 > \left(\frac 5 2\right)^n=6.25$$ $$n=3\implies a_n=20 > \left(\frac 5 2\right)^n=15.625$$ $$n=4\implies a_n=47 > \left(\frac 5 2\right)^n=39.0625$$ $$n=5\implies a_n=114 > \left(\frac 5 2\right)^n=97.65625$$ $$n=6\implies a_n=275 > \left(\frac 5 2\right)^n=244.140625$$ $$n=7\implies a_n=664 > \left(\frac 5 2\right)^n=610.3515625$$ $$n=8\implies a_n=1603 > \left(\frac 5 2\right)^n=1525.87890625$$ $$n=9\implies a_n=3870 > \left(\frac 5 2\right)^n=3814.697265625$$ $$n=10\implies a_n=9343 \leq \left(\frac 5 2\right)^n=9536.7431640625$$

Thus, your value of $n_0$ is $10$ since that's when $a_n \leq \left(\frac 5 2\right)^n$ first becomes true.