Question

an + an-1 - 10an-2 + 8an-3

Answer 1

Since, given question is incomplete, so we assume as follows:

Consider the recurrence relation $a_{n}=a_{n-1}+6 . a_{n-2}, \quad n \geq 2, a_{0}=3, a_{1}=6$ The characteristic equation is $r^{2}-r-6=0$ .

$\begin{array}{r} r^{2}-r-6=0 \\ r^{2}-3 r+2 r-6=0 \\ r(r-3)+2(r-3)=0 \\ (r-3)(r+2)=0 \\ r=3,-2 \end{array}$

The roots of the characteristic equation are r_{1}=3, r_{2}=-2.

For n=0 the solution of the recurrence relation is,

$\begin{array}{ll} a_{0} & =\alpha_{1} r_{1}^{0}+\alpha_{2} r_{2}^{0} \\ 3 & =\alpha_{1}(3)^{0}+\alpha_{2}(-2)^{0} \\ 3 & =\alpha_{1}+\alpha_{2} \ ...(1) \end{array} \quad \text { As } a_{0}=3$

For n=1 the solution of the recurrence relation is,

$\begin{array}{ll} a_{1} & =\alpha_{1} \cdot\left(r_{1}\right)^{1}+\alpha_{2}\left(r_{2}\right)^{1} \\ 6 & =\alpha_{1} \cdot(3)^{1}+\alpha_{2}(-2)^{1} \\ 6 & =3 \alpha_{1}-2 \alpha_{2}\ ...(2) \end{array} \quad \text { As } a_{1}=6$

Solve equation (1) and (2) to find the values of $\alpha_{1}, \alpha_{2}$ .

Multiply equation (1) by 2 and add equation (1) and (2).

$\begin{aligned} 6 &=2 \alpha_{1}+2 \alpha_{2} \\ 6 &=3 \alpha_{1}-2 \alpha_{2} \\ 12 &=5 \alpha_{1} \\ \alpha_{1} &=\frac{12}{5} \end{aligned}$

Thus, the value of $\alpha_{2}$ is,

$\begin{array}{r} \frac{12}{5}+\alpha_{2}=3 \\ \alpha_{2}=\frac{3}{5} \end{array}$

Therefore, the solution of the recurrence relation is,

$\begin{aligned} a_{n} &=\alpha_{1} r_{1}^{n}+\alpha_{2} r_{2}^{n} \\ &=\left(\frac{12}{5}\right)(3)^{n}+\left(\frac{3}{5}\right)(-2)^{n} \end{aligned}$