Question

Find the general term of sequence defined by these conditions.

Answer 1

Answer:

$\displaystyle a_{n} = (2)^{2n -1} - (3) ^{n-1 }$

Step-by-step explanation:

we want to figure out the general term of the following recurrence relation

$\displaystyle \rm a_{n + 2} - 7a_{n + 1} + 12a_n = 0 \: \: where : \: \:a_1 = 1 \: ,a_2 = 5,$

we are given a linear homogeneous recurrence relation which degree is 2. In order to find the general term ,we need to make it a characteristic equation i.e

• ${x}^{n} = c_{1} {x}^{n - 1} + c_{2} {x}^{n - 2} + c_{3} {x}^{n -3 } { \dots} + c_{k} {x}^{n - k}$

the steps for solving a linear homogeneous recurrence relation are as follows:

1. Create the characteristic equation by moving every term to the left-hand side, set equal to zero.
2. Solve the polynomial by factoring or the quadratic formula.
3. Determine the form for each solution: distinct roots, repeated roots, or complex roots.
4. Use initial conditions to find coefficients using systems of equations or matrices.

Step-1:Create the characteristic equation

${x}^{2} - 7x+ 12= 0$

Step-2:Solve the polynomial by factoring

factor the quadratic:

$( {x}^{} - 4)(x - 3) = 0$

solve for x:

$x = \rm 4 \:and \: 3$

Step-3:Determine the form for each solution

since we've two distinct roots,we'd utilize the following formula:

$\displaystyle a_{n} = c_{1} {x} _{1} ^{n } + c_{2} {x} _{2} ^{n }$

so substitute the roots we got:

$\displaystyle a_{n} = c_{1} (4)^{n } + c_{2} (3) ^{n }$

Step-4:Use initial conditions to find coefficients using systems of equations

create the system of equation:

$\begin{cases}\displaystyle 4c_{1} +3 c_{2} = 1 \\ 16c_{1} + 9c_{2} = 5\end{cases}$

solve the system of equation which yields:

$\displaystyle c_{1} = \frac{1}{2} \\ c_{2} = - \frac{1}{3}$

finally substitute:

$\displaystyle a_{n} = \frac{1}{2} (4)^{n } - \frac{1}{3} (3) ^{n }$

$\displaystyle \boxed{ a_{n} = (2)^{2n-1 } - (3) ^{n -1}}$

and we're done!