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.
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.
$$7r_{n-1}-10r_{n-2}=7(3.2^{n-1}-4.5^{n-1})-10(3.2^{n-2}-4.5^{n-2})$$ $$7r_{n-1}-10r_{n-2}=(42-30).2^{n-2}+(-140+2=40)5^{n-2}$$ $$7r_{n-1}-10r_{n-2}=3.2^n-4.5^n$$ $$7r_{n-1}-10r_{n-2}=r_n$$
Based on the input I was given by some members here's what I came up with following some of those steps and putting it in my own words in a way to make sure I understand what's happening:
$r_n = 3\cdot2^n - 4\cdot5^n$, $n \geq 0$
$r_n = 7r_{n-1} - 10r_{n-2}$, $n \geq 2$
$n = 1$
$1-1 = 1$
Working the left side first
$7\left(3\cdot2^1 - 4\cdot5^{1} \right)$
$7\left(\frac{3}{2}\cdot2 - \frac{4}{5}\cdot5 \right)$
$7\left(\left(1.5\cdot2\right) -\left(0.8\cdot5\right)\right)$
3, -4 <-- These are the first numbers in 3*2 and -4*5 (but ignoring -4 for now)
Moving the 3 the to the outside now. Repeat the process for the 7.
$3\left(7\cdot2^1 - 10\cdot2^{1} \right)$
$3\left(\frac{7}{2}\cdot2 - \frac{10}{4}\cdot2 \right)$ <-- Multiply the denominator by $2^n$ for the denominoator of the second part.
$3\left(\left(3.5\cdot2\right) -\left(2.5\cdot2\right)\right)$
$\left(7-5\right)$
$2$ <-- Bring down the 3
$3\cdot2$ <-- These are the first numbers in $r_n$
Now I have 3, 2 or $3 \cdot 2$ of $r_n$
Working the right side now
$-10\left(3\cdot2 - 4\cdot5 \right)$
$-10\left(\frac{3}{2}\cdot2 - \frac{4}{5}\cdot5 \right)$
$-10\left(\left(1.5\cdot2\right) -\left(0.8\cdot5\right)\right)$
3, -4 <-- We have the same values again this time we use the -4 and ignore the 3
Moving the -4 to the outside and repeating the same process above when we moved the 3 to the outside and moved the 7 inside.
$-4\left(7\cdot5 - 10\cdot5 \right)$
$-4\left(\frac{7}{5}\cdot5 - \frac{10}{25}\cdot5 \right)$ <-- Multiply the denominator by $5^n$ for the denominoator of the second part.
$-4\left(\left(1.4\cdot5\right) -\left(0.4\cdot5\right)\right)$
$\left(7-2\right)$
$5$ <-- Bring down the -4
$-4\cdot5$ <-- These are the first numbers in $r_n$
Now I have -4, 5 or $-4 \cdot 5$ of $r_n$
We then have: $r_n = 3\cdot2^n - 4\cdot5^n$, $n \geq 0$
