Find the value ${a_{10}}^2+4a_{9}-a_9a_{11}$

Question

Let $a_0=3$ and $a_1=7$ and $a_{n+2}-5a_{n+1}+6a_n=2$ for all $n$. What's the value of $${(a_{10})}^2+4a_{9}-a_9a_{11}$$

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I have successfully solved the recursion and obtained the explicit formula $a_n=2\times 3^n+1$. However, I am currently facing challenges in progressing further.

Answer 1

$$ \left( \begin{array}{cccccc} n & 3^n & a_n & a_n^2 - a_{n+1} a_{n-1} & 4 a_{n-1} & add \\ 0 & 1 & 3 & & & \\ 1 & 3 & 7 & -8 & 12 & 4 \\ 2 & 9 & 19 & -24 & 28 & 4 \\ 3 & 27 & 55 & & & \\ 4 & 81 & 163 & & & \\ 5 & 243 & 487 & & & \\ \end{array} \right) $$ You should fill this table in. the thing with the square, $7^2 - 19 \cdot 3 = 49 - 57 =-8.$ Next $ 19^2 - 55 \cdot 7 = 361 - 385 = -24.$ So, what is $55^2 - 163 \cdot 19 \; ? \; \; $

the predictable thing is that $a_n^2 - a_{n+1} a_{n-1} $ is fairly simple because of cancellation. Just write it out, $$ a_{n+1} a_{n-1} - a_n^2 = 8 \cdot 3^{n-1} $$ so $$ a_n^2 - a_{n+1} a_{n-1} = -8 \cdot 3^{n-1} $$ and $$ a_{n-1} = 1 + 2 \cdot 3^{n-1}.$$

Multiply by $4$ and add, you get $4.$

Answer 2

Let $x=3^9$ and simplify, which I think can reasonably be done within a minute if one is familiar with binomial expansions or the formulas for the coefficients.

$$\begin{align*} A &= a_{10}^2 + 4a_9 - a_9 a_{11} \\ &= \left(2\cdot3x + 1\right)^2 + 4\left(2x + 1\right) - \left(2x + 1\right) \left(2\cdot9x + 1\right) \\ &= 36 \left(x + \frac16\right)^2 + 8\left(x + \frac12\right) - 36\left(x + \frac12\right) \left(x + \frac1{18}\right) \\ &\,\,\vdots \end{align*}$$

but note that $A\neq7$.