Recipe

Solve a linear recurrence with a repeated characteristic root

Use when: a second-order linear recurrence's characteristic equation factors as a perfect square, giving one repeated root instead of two distinct roots

Goal: the closed form $a_n=(A+Bn)r_0^n$ , with $A$ and $B$ correctly pinned down by the initial conditions

Taught in Repeated characteristic roots

Quick recipe

Solve the characteristic equation and confirm it's a repeated root
Write the general solution as $(A+Bn)r_0^n$ , not $Ar_0^n$ alone
Substitute the initial conditions and solve for $A$ and $B$
Write the final closed form and check it against the recurrence

1
Form the characteristic equation from the recurrence and factor it; a repeated root shows up as a perfect square.

Example

For $a_n=8a_{n-1}-16a_{n-2}$ : $r^2-8r+16=(r-4)^2=0$ , a double root at $r_0=4$ .

Why: the general solution's shape depends entirely on whether the root repeats; a distinct-root solution form would be missing a genuine second solution here.
2
Multiply the second copy of $r_0^n$ by $n$ to get a genuinely independent second solution.

Example

$a_n=(A+Bn)4^n$ .

Why: two copies of the same building block $r_0^n$ can't match two separate initial conditions; the extra factor of $n$ is what supplies the missing, genuinely different solution.
3
Plug in the two given initial values to build two equations in $A$ and $B$ , then solve them.

Example

$a_0=2$ gives $A=2$ . $a_1=12$ gives $(2+B)\cdot 4=12$ , so $B=1$ .

Why: two initial conditions are exactly enough to pin down the two unknown constants in this two-parameter family.
4
State $a_n$ with the found constants, then verify it reproduces a value computed directly from the original recurrence.

Example

$a_n=(2+n)4^n$ . Check $a_2$ : the closed form gives $(2+2)\cdot16=64$ ; the recurrence gives $8(12)-16(2)=64$ , matching.

Why: this catches an arithmetic slip in solving for $A$ or $B$ before the closed form gets used further downstream.

Common mistakes

Writing $a_n=Ar_0^n+Br_0^n$ for a repeated root; that's the same function counted twice, not two independent solutions.
Forgetting the factor of $n$ entirely and trying to force a one-parameter family to match two initial conditions.
Using $a_0$ to solve for $B$ or $a_1$ to solve for $A$ in the wrong order; substituting $n=0$ correctly leaves only $A$ , and only $a_1$ 's equation involves $B$ as well.
Confusing a repeated root with two very close but genuinely distinct roots; the discriminant must be checked to be exactly zero, not just small.