Find a closed form for the recursive sequence $a_0=2, a_1=6$

Key concept: This recurrence is the kind that rewards a clean characteristic-equation setup. Once you solve the polynomial, the sequence falls into place quickly.

Step by step

We have

$$a_0=2,\quad a_1=6,\quad a_k=2a_{k-1}+3a_{k-2}\ (k\ge 2).$$

To find a closed form, try a solution of the form

$a_k=r^k.$

Substitute into the recurrence:

$r^k=2r^{k-1}+3r^{k-2}.$

Divide by $r^{k-2}$:

$r^2=2r+3.$

So the characteristic equation is

$r^2-2r-3=0,$

which factors as

$(r-3)(r+1)=0.$

Thus the roots are $r=3$ and $r=-1$.

So the general form is

$a_k=A\cdot 3^k+B\cdot(-1)^k.$

Use the initial conditions:

For $k=0$:

$A+B=2.$

For $k=1$:

$3A-B=6.$

Add the equations:

$4A=8 \Rightarrow A=2.$

Then

$B=0.$

So the sequence is simply

$\boxed{a_k=2\cdot 3^k}.$

You can check it quickly:

• $a_0=2\cdot 3^0=2$
• $a_1=2\cdot 3^1=6$
• $a_2=2\cdot 9=18$, and the recurrence gives $2(6)+3(2)=18$.

That matches perfectly.

Pitfall alert

A frequent slip is to stop after finding the roots and forget to use both initial values. Another one: writing $A3^k+B1^k$ instead of $A3^k+B(-1)^k$. The sign matters, especially when checking odd terms.

Try different conditions

If the initial conditions changed, say $a_0=1$ and $a_1=7$, the same characteristic equation would still apply, but the constants $A$ and $B$ would change. The method stays the same: solve the recurrence polynomial first, then fit the starting values. If the recurrence were non-homogeneous, such as $a_k=2a_{k-1}+3a_{k-2}+5$, you would need an extra particular solution.

Further reading

characteristic equation, linear recurrence, closed form