Formula for Shifted Geometric Squence

Question

In my math class, we are currently studying recursive formulas like the following:

Geometric sequence:

$U_0 =$ start
$U_n =$ ratio$ \cdot U_{n-1}$

Arithmetic sequence:

$U_0 =$ start
$U_n = U_{n-1} + $difference

Shifted Geometric sequence:

$U_0 =$ start
$U_n = $ratio$ \cdot U_{n-1} + $difference

I know that a Geometric sequence can be modeled by this:

$Y =$ start(ratio)$^X$

I know that a Arithmetic sequence can be modeled by this:

$Y =$ difference$\cdot X + $start

How do you model a Shifted Geometric sequence in the second listed format?

Answer 1

Let's look at the sequence with initial value $U_0$ and the rule $U_n = AU_{n-1}+B$. We can look at some example terms:

$$U_0 = U_0$$

$$U_1 = A(U_0 + B) + B = AU_0 + AB + B$$

$$U_2 = A(AU_0 + AB + B) + B = A^2U_0 + A^2B + AB + B$$

Now we can see that $U_n = A^nU_0 + B(1+A+A^2+...+A^n)$. The geometric series $1+A+A^2+...+A^n=n+1$ when $A=1$ and $1+A+A^2+...+A^n=\frac{1-A^{n+1}}{1-A}$ when $A\ne1$. Now, we can put it all together and say $U_n=U_0+B(n+1)$ when $A = 1$ and $U_n = A^nU_0+B\frac{1-A^{n+1}}{1-A}$ if $A\ne1$.

Answer 2

$$\begin{align} U_n & =rU_n+\Delta \\ & =r(rU_{n-2}+\Delta)+\Delta \\ & =r^2U_{n-2}+\Delta+r\Delta \\ & =r^3U_{n-3}+\Delta+r\Delta+r^2\Delta \\ \vdots & \qquad\vdots \\ U_n & = r^nU_0+\Delta(r^0+r^1+r^2+\dots r^n) \\ \end{align}$$

And so:

$$U_n = r^nU_0+\Delta\frac{1-r^n}{1-r}$$

assuming $r\ne1$, in which case it would reduce down to an arithmetic sequence.

Answer 3

$u_0\\ u_1 = r u_0 + d\\ u_2 = r u_2 + d = r^2 u_0 + r d + d\\ u_3 = r^3 u_0 + (r^2 + r + 1) d$

Have you seen $(r^2 + r + 1)$ before?

This is the geometric series,which is the partial sum of a geometric sequence.

$(r^2 + r + 1) = \sum_\limits {i=0}^2 r^i = \frac {(r^3-1)}{r-1}$ and to abstract it even further.

$(r^2 + r + 1) = \sum_\limits {i=0}^{n-1} r^i = \frac {(r^{n}-1)}{r-1}$

$u_n = r^n u_0 + d\frac {(r^{n}-1)}{r-1}$