1 , 35, 156,?

asked by guest
on Nov 25, 2024 at 9:23 am



MathBot Answer:

$$1,35,156,...$$The next term in this recurrent sequence is $a_{4}=\frac{19945}{34}$.


The nth term in this sequence is given by the formula:

Explicit Formula: $$a_n=\frac{39304 \left(\frac{121}{34}\right)^{n}}{10527} - \frac{1069}{87}$$

Recursive Formula: $$a_n=\frac{121 a_{n-1}}{34} + \frac{1069}{34}, \text{where } a_{1}=1$$


RECURSIVE FORMULA

Linear Recurrence Relation

[View Steps]

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 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 ]

To find $a_p$, find an appropriate trial solution. [ show ]

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 $c_i$: $$\begin{aligned}a_1 &= c_1 f_1(1) + c_2 f_2(1) + ... + c_k f_k(1) + t(1) \\ a_2 &= c_1 f_1(2) + c_2 f_2(2) + ... + c_k f_k(2) + t(2) \\ \vdots \\ a_k &= c_1 f_1(k) + c_2 f_2(k) + ... + c_k f_k(k) + t(k) \end{aligned}$$