Prove that $a_n-10a_{n-1}+a_{n-2}=0$.

Question

Let $a_n = (5+2\sqrt{6})^n+(5-2\sqrt{6})^n$. Prove that $a_n-10a_{n-1}+a_{n-2}=0$.

I think this depends on whether $n$ is even or odd so in the case $n$ is even we have $a_n = 2(\binom{n}{0}5^n+\binom{n}{2}5^{n-2}(2\sqrt{6})^2+\cdots+\binom{n}{n}(2\sqrt{6})^n)$, but it seems computational to do it this way. Is there an easier way?

Answer 1

hint: You can solve backward by noting that: $a_0 = 2, a_1 = 10$, then you have $x^2-10x+1 = 0$. What are the roots $r_1, r_2$ for this equation, and then $a_n = Ar_1^n + Br_2^n$. You can solve for $A,B$ and it should match with your formula above.

Answer 2

If $a_n = C \alpha^n + D \beta^n $ with $\alpha\neq\beta$, $\alpha$ and $\beta$ are roots of the characteristic polynomial $$ x^2 - Ux - V$$ that gives $a_{n+2}=U a_{n+1}+ V a_n$. By Viète's theorem, $U=\alpha+\beta$ and $V=-\alpha\beta$.
In our case, $U=10$ and $V=-1$ by straightforward computations.

Answer 3

Just do it directly: $$a_n-10a_{n-1}+a_{n-2}$$ $$=(5+2\sqrt{6})^n+(5-2\sqrt{6})^n-10[(5+2\sqrt{6})^{n-1}+(5-2\sqrt{6})^{n-1}]+(5+2\sqrt{6})^{n-2}+(5-2\sqrt{6})^{n-2}$$ $$=(5+2\sqrt{6})^{n-2}\left([5+2\sqrt{6}]^2-10[5+2\sqrt{6}]+1\right)+(5-2\sqrt{6})^{n-2}\left([5-2\sqrt{6}]^2-10[5-2\sqrt{6}]+1\right)\qquad (1)$$ Now consider the polynomial $$x^2-10x+1\qquad (2)$$ It has roots $$x=5\pm2\sqrt{6}$$ And if you look closely, in $(1)$ both $(5+2\sqrt{6})^{n-2}$ and $(5-2\sqrt{6})^{n-2}$ are being multiplied by $(2)$ evaluated at it's roots making their coefficients $0$ and the whole expression $0$.

This is similar to what Jack D'Aurizio mentioned but noticing the theorem in work in this specific case instead of skipping directly to applying the conclusion of said theorem.