Question

Find the solution to the recurrence relation a_n=〖-3a〗_(n-1) 〖-3a〗_(n-2) 〖-a〗_(n-3) with initial conditions a_0=1,a_1=-2 and a_2=-1

Answer 1

We have recurrence relation

$$a _ {n} = - 3 a _ {n - 1} - 3 a _ {n - 2} - a _ {n - 3}$$

Characteristic polynomial:

$$\lambda^ {3} + 3 \lambda^ {2} + 3 \lambda + 1 = 0$$ $$(\lambda + 1) ^ {3} = 0$$ $$\lambda = - 1$$

SO we have one root $\lambda = -1$ with the $3^{\mathrm{rd}}$ multiplicity.

So the solution of the relation is

$$a _ {n} = (- 1) ^ {n} (c _ {1} + c _ {2} n + c _ {3} n ^ {2})$$

Substituting initial conditions we get:

$$a _ {0} = c _ {1} = 1$$ $$a _ {1} = - 1 \left(c _ {1} + c _ {2} + c _ {3}\right) = - 2$$ $$a _ {2} = c _ {1} + 2 c _ {2} + 4 c _ {3} = - 1$$

Substituting $c_{1} = 1$ into the $2^{\text{nd}}$ and $3^{\text{rd}}$ equations we get:

$$c _ {2} + c _ {3} = 1$$ $$c _ {2} + 2 c _ {3} = - 1$$

Solving this system we get:

$$c _ {1} = 1$$ $$c _ {2} = 3$$ $$c _ {3} = - 2$$

Thus the solution of the recurrence relation is

$$a _ {n} = (- 1) ^ {n} (1 + 3 n - 2 n ^ {2})$$