$$\frac{6}{5},\frac{4}{3},\frac{3}{2},...$$The 10th term in this recurrent sequence is $\frac{456161}{98304}$.
The nth term in this sequence is given by the formula:
Explicit Formula: $$a_n=\frac{32 \left(\frac{5}{4}\right)^{n}}{75} + \frac{2}{3}$$
Recursive Formula: $$a_n=\frac{5 a_{n-1}}{4} - \frac{1}{6}, \text{where } a_{1}=\frac{6}{5}$$
RECURSIVE FORMULA
Linear Recurrence Relation
[View Steps]
Given a sequence of 3 terms, the recursive formula is of the form $$a_n=x_0 + x_1 a_{n-1} + ... + x_k a_{n-k}, \text{where } 1 \leq k \leq 1$$
When $k=1$: $$a_n=x_0 + a_{n-1} x_{1}, n > 1$$ $$\begin{aligned}\frac{4}{3}&=x_{0} + \frac{6 x_{1}}{5}\\\frac{3}{2}&=x_{0} + \frac{4 x_{1}}{3} \end{aligned}$$$$\left\{ x_{0} : - \frac{1}{6}, \ x_{1} : \frac{5}{4}\right\}$$
The nth term rule is:$$a_n=\frac{5 a_{n-1}}{4} - \frac{1}{6}$$
Given a sequence of m terms, the recursive formula is of the form $$a_n=x_0 + x_1 a_{n-1} + ... + x_k a_{n-k}, \text{where } 1 \leq k \leq \left \lfloor \frac{m-1}{2} \right \rfloor$$
Using all the given terms, solve the systems of equations for $x_i$ when $k=1,...,\left \lfloor \frac{m-1}{2} \right \rfloor$. If $x_i$ is not found for any $k$, a recursive formula cannot be found using this method.
When $k=1$: $$a_n=x_0 + x_1 a_{n-1}, n > 1$$ Solve for $x_0$ and $x_1$: $$\begin{aligned} a_2&=x_0 + x_1 a_1 \\ a_3&=x_0 + x_1 a_2 \\ \vdots \\ a_m&=x_0 + x_1 a_{m-1}\end{aligned}$$
EXPLICIT FORMULA
Non-Homogeneous Solution
[View Steps]
Given a non-homogeneous linear recurrence relation $$a_n=\frac{5 a_{n-1}}{4} - \frac{1}{6}$$ the closed formula where $a_h$ is the general solution of $\frac{5 a_{n-1}}{4}$, and $a_p$ is the particular solution of $- \frac{1}{6}$ is $$a_n=a_h + a_p$$
Find $a_h$:
characteristic equation: $r - \frac{5}{4}=0$
roots: $\left\{ \frac{5}{4} : 1\right\}, \text{ where r:m}$
general solution: $a_h=\left(\frac{5}{4}\right)^{n} c_{1}$
Find $a_p$:
trial solution: $t(n)=A$
particular solution: $a_p=\frac{2}{3}$
$\begin{aligned}A&=\frac{5 A}{4} - \frac{1}{6} \\ A&=\frac{2}{3}\end{aligned}$
Using the first $1$ terms, solve for $c_i$: $$a_n=\left(\frac{5}{4}\right)^{n} c_{1} + \frac{2}{3}$$ $$\begin{aligned}\frac{6}{5}&=\left(\frac{5}{4}\right)^{1} c_{1} + \frac{2}{3}\end{aligned}$$ $$\left\{ c_{1} : \frac{32}{75}\right\}$$
The nth term rule is: $$a_n=\frac{32 \left(\frac{5}{4}\right)^{n}}{75} + \frac{2}{3}$$
Given a non-homogeneous linear recurrence relation in the form $$a_n=C_1 a_{n-1} + C_2 a_{n-2} + ... + C_k a_{n-k} + f(n), \text{where } f(n)\neq0$$the closed formula where $a_h$ is the general solution of $C_1 a_{n-1} + C_2 a_{n-2} + ... + C_k a_{n-k}$ (homogeneous) and $a_p$ is the particular solution of $f(n)$ is $$a_n=a_h + a_p$$
To find $a_h$, use the Characteristic Root Technique. [ show ]
Given a homogeneous linear recurrence relation of order $k$$$C_1 a_{n-1} + C_2 a_{n-2} + ... + C_k a_{n-k}$$ the characteristic equation is the $k$th degree polynomial equation$$r^k - C_1 r^{k-1} - C_2 r^{k-2} - ... - C_{k-1} r - C_k=0$$
Characteristic Roots
1. If $r$ is a distinct real root, then the general solution includes a term $r^n$.
2. If $r$ is a repeated real root with multiplicity of $m$, then the general solution includes the terms $r^n, n r^n,..., n^{m-1} r^n$.
3. If $r$ is a pair of complex roots, then the general solution includes the terms $|r|^n \sin(n\theta)$ and $|r|^n \cos(n\theta)$ where $r=a\pm b i= |r|e^{\pm \theta i}$.
Let the terms be denoted by $f_i(n)$ where $i=1,...,k$. The general solution is $$a_h=c_1 f_1(n) + c_2 f_2(n) + ... + c_k f_k(n)$$
To find $a_p$, find an appropriate trial solution. [ show ]
1. Choose a reasonable trial solution $t(n)$ for $f(n)$.
Possible Trial Solutions for f(n)
1. Constant
A
2. mth Degree Polynomial
Amnm + ... + A1n + A0
3. Exponential with base b
Abn OR Anbn OR etc.
2. Substitute $a_i = t(i)$ for $i = n, n-1, ..., n-k$ in the recurrence relation, and solve for the missing variables $A_j$.
3. The particular solution is $$a_p=t(n)$$
The closed formula where $c_i$ are coefficients determined by the initial conditions is $$a_n=c_1 f_1(n) + c_2 f_2(n) + ... + c_k f_k(n) + t(n)$$
Solve the systems of equations to find $c_i$: $$\begin{aligned}a_1 &= c_1 f_1(1) + c_2 f_2(1) + ... + c_k f_k(1) + t(1) \\ a_2 &= c_1 f_1(2) + c_2 f_2(2) + ... + c_k f_k(2) + t(2) \\ \vdots \\ a_k &= c_1 f_1(k) + c_2 f_2(k) + ... + c_k f_k(k) + t(k) \end{aligned}$$
