Please help me with this linear recurrence question question. i have spent the past two hours on it with no success. It will no doubt be easy for someone but unfortunately not for me!
I need to find a closed form for
$$X_{n+1} =0.9X_n −5,\quad X_1= 20.$$
HINT: There are many ways to solve this particular type of recurrence; here’s one. Any recurrence of the form $x_{n+1}=ax_n+b$ can be reduced to the form $y_{n+1}=cy_n$ by a change of variable. Specifically, let $y_n=x_n-d$ for some constant $d$ that will be determined later. Then $x_n=y_n+d$, and your recurrence $x_{n+1}=0.9x_n-5$ becomes $y_{n+1}+d=0.9(y_n+d)-5$, or
$$y_{n+1}=0.9y_n-5-0.1d\;.\tag{1}$$
Setting $-5-0.1d=0$ and solving for $d$, we get $d=-50$; substituting $d=-50$ into $(1)$ leaves us with the very simple recurrence
$$y_{n+1}=0.9y_n\;,\tag{2}$$
with initial value $y_1=x_1-d=20-(-50)=70$.
Now solve $(2)$ to get a closed form for $y_n$, and use the relationship $x_n=y_n+d=y_n-50$ to get a closed form for $x_n$.
A bit more generally, for
$$X_{n+1} = aX_n + b,$$
the closed form is
$$X_n = \left( \sum_{k=0}^{n-2} a^k \right) b + a^{n-1} X_1 = \frac{a^{n-1}-1}{a-1}b + a^{n-1} X_1.$$
A proof by induction should be trivial.
