Problem 890

The characteristic equation of the recurrence relation
$$a_{n}=5 a_{n-1}-6 a_{n-2} \quad(n \geqslant 2)$$

is $x^{2}-5 x+6=0$, with characteristic roots $x_{1}=2, x_{2}=3$. Therefore, the general solution is
$$a_{n}=c_{1} \cdot 2^{n}+c_{2} \cdot 3^{n} .$$

Since (3.7) does not have a characteristic root equal to 1, the recurrence relation
$$a_{n}=5 a_{n-1}-6 a_{n-2}+n+2 \quad(n \geqslant 2)$$

has a particular solution $a_{n}=A n+B$, where $A$ and $B$ are constants to be determined. Substituting into (3.8) yields
$$\begin{aligned} A n+B= & 5[A(n-1)+B]- \\ & 6[A(n-2)+B]+n+2, \end{aligned}$$

simplifying to
$$2 A n+2 B-7 A=n+2,$$

so
$$\left\{\begin{array}{l} 2 A=1 \\ 2 B-7 A=2 \end{array}\right.$$

Solving these, we get $A=\frac{1}{2}, B=\frac{11}{4}$. Thus,
$$a_{n}=c_{1} \cdot 2^{n}+c_{2} \cdot 3^{n}+\frac{1}{2} n+\frac{11}{4},$$

where $c_{1}, c_{2}$ are constants to be determined. Using the initial conditions, we have
$$\left\{\begin{array}{l} c_{1}+c_{2}+\frac{11}{4}=\frac{27}{4}, \\ 2 c_{1}+3 c_{2}+\frac{1}{2}+\frac{11}{4}=\frac{49}{4} . \end{array}\right.$$

Solving these, we get $c_{1}=3, c_{2}=1$. Therefore,
$$a_{n}=3 \cdot 2^{n}+3^{n}+\frac{1}{2} n+\frac{11}{4} \quad(n \geqslant 0) .$$