I have two recursive formulas that I don’t know how to convert to explicit form.
$$ x_1 = 3 $$ $$ x_n = x_{n-1} + 6*4^{n-2} $$ $$ y_1 = 6 $$ $$ y_n = y_{n-1} + 15*4^{n-2} $$
Since they follow the same format, instruction for converting either one should be enough enough for me to figure out the other. How do I convert?
Any method that finds a valid solution is appreciated, regardless of approach.
Note that $x_{1}=3, x_{2}=9$. Then, using the recursive definition, we have: $$x_{n} = x_{n-1} + 6\cdot 4^{n-2}$$ $$x_{n+1} = x_{n}+6\cdot 4^{n-1}$$ Multiplying both sides of the first equation by $4$, and then subtracting it from the second equation: $$x_{n+1}-4x_{n}=(x_{n}+6\cdot 4^{n-1})-4(x_{n-1}+6\cdot 4^{n-2})$$ $$x_{n+1}-5x_{n}+4x_{n-1}=0$$ Now, note that the characteristic polynomial for this recurrence is $r^{2}-5r+4$, with roots $r=1$ and $r=4$. Thus, we must have, for some constants $a$ and $b$: $$x_{n}=a\cdot 4^{n}+b\cdot 1^{n}$$ Using the initial conditions: $$x_{1}=3\Rightarrow 4a + b = 3$$ $$x_{2}=9\Rightarrow 16a + b = 9$$ Solving the system of equations, we find $\displaystyle(a,b) = \bigg(\frac{1}{2}, 1\bigg)$. Thus, $\displaystyle\boxed{x_{n}=\frac{4^{n}}{2}+1.}$
You can solve the second recurrence similarly.