<s>[INST] Prove that {$r_n$} satisfies $r_n = 7r_{n-1} - 10r_{n-2}, n \geq 2$ Problem: For the sequence $r$ defined by 
$$r_n = 3 \cdot 2^n - 4 \cdot 5^n, \ \ \ n \geq 0$$ 
Prove that {$r_n$} satisfies 
$$r_n = 7r_{n-1} - 10r_{n-2}, \ \ \ n \geq 2$$
Can this problem be explained and broken down and show the process? I'd like to follow your steps on my own.
I've been studying this question but the opening multiplication is throwing me off. 
Similar Problem
 [/INST] To show that $r_n = 3\cdot 2^n  - 4\cdot 5^n$ satisfy the equation then start by writing the equation as
$$r_n - 7r_{n-1} + 10r_{n-2} = 0$$
Now calculate $r_n, r_{n-1}$ and $r_{n-2}$ from the formula. This gives us
$$r_n = \color{red}{3\cdot 2^n  - 4\cdot 5^n}$$
$$r_{n-1} = 3\cdot 2^{n-1}  - 4\cdot 5^{n-1} = \color{blue}{\frac{3}{2}\cdot 2^{n}  - \frac{4}{5}\cdot 5^{n}}$$
$$r_{n-2} = 3\cdot 2^{n-2}  - 4\cdot 5^{n-2} = \color{green}{\frac{3}{2^2}\cdot 2^{n}  - \frac{4}{5^2}\cdot 5^{n}}$$
where we have used the rule $x^{a+b} = x^a \cdot x^b$ and $x^{-a} = \frac{1}{x^a}$ to simplify (for example $2^{n-2} = 2^{n}\cdot 2^{-2} =2^n\cdot \frac{1}{4}$). Now we substitute this into the equation $r_n - 7r_{n-1} + 10r_{n-2} = 0$ to find
$$\left(\color{red}{3\cdot 2^n  - 4\cdot 5^n}\right) - 7\cdot\left(\color{blue}{\frac{3}{2}\cdot 2^{n}  - \frac{4}{5}\cdot 5^{n}}\right) + 10 \cdot \left(\color{green}{\frac{3}{2^2}\cdot 2^{n}  - \frac{4}{5^2}\cdot 5^{n}}\right) = 0$$
and after rearranging we get
$$\left(\color{red}{3\cdot 2^n} - 7\cdot \color{blue}{\frac{3}{2}\cdot 2^n} + 10\cdot\color{green}{\frac{3}{2^2}\cdot 2^n }\right) + \left(-\color{red}{4\cdot 5^n} + 7\cdot \color{blue}{\frac{4}{5}\cdot 5^n} - 10\cdot\color{green}{\frac{4}{5^2}\cdot 5^n }\right) = 0$$
Now we take the $2^n$ and $5^n$ outside of the brackets to find
$$2^n\cdot\left(\color{red}{3} - 7\cdot \color{blue}{\frac{3}{2}} + 10\cdot\color{green}{\frac{3}{2^2}}\right) + 5^n\cdot \left(-\color{red}{4} + 7\cdot \color{blue}{\frac{4}{5}} - 10\cdot\color{green}{\frac{4}{5^2}}\right) = 0$$
By calculating he sums in the brackets we find that they are both $0$ so we are left with $0=0$ which shows that the formula for $r_n$ does indeed satify the equation.
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