Question

an=3an−1+n2−3,n≥,a0=1

Answer 1

$$a_n=3a_{n-1}+n^2-3, n\geq0, a_0=1$$

The homogeneous part of this recurrence relation is;

$$a_n=3a_{n-1}$$

The characteristics equation for this is;

$$r=3$$

Thus, the solution for the homogeneous part is;

$$a_n=\alpha3^n$$

The complementary solution is:

$$a_n^{(c)}= \alpha 3^n$$

The non homogeneous part of the relation is $n^2-3$

In this case, the particular solution is of the form:

$$a_n^{(p)}=p_2n^2+p_1n+p_0$$

Substituting into the original relation, we have that:

$$p_2n^2+p_1n+p_0=3(p_2(n-1)^2+p_1(n-1)+p_0)+n^2-3\\ p_2(-2n^2+6n-3)+p_1(-2n+3)-2p_0=n^2-3\\ -2p_2n^2+(6p_2-2p_1)n+(-3p_2+3p_1-2p_0)=n^2-3\\$$

We now have that;

$-2p_2=1 \implies p_2=-\frac{1}{2}\\ 6p_2-2p_1=0 \implies p_1=3p_2 \implies p_1=-\frac{3}{2}\\ -3p_2+3p_1-2p_0=-3 \implies -2p_0=0 \implies p_0=0$

Thus the particular solution is

$$a_n^{(p)}=-\frac{1}{2}n^2-\frac{3}{2}n$$

Hence, the general solution which is $a_n^{(c)}+a_n^{(p)}$ is:

$$a_n=\alpha 3^n-\frac{1}{2}n^2-\frac{3}{2}n$$

Substitute $a_0=1$ to get the value of $\alpha$ . We have that;

$$a_0=\alpha 3^0 \implies \alpha=1$$

Hence the general solution of the recurrence relation is:

$$a_n=3^n-\frac{1}{2}n^2-\frac{3}{2}n, n\geq 0$$