A sequence satisfies a(n) = 4a(n-1) - 4a(n-2) for every integer n at least 2, with a(0) = -2025 and a(1) = 2026. Given that a(2025) = b times 2^c for constants b and c, find b/c.

Step-by-step solution

1. 1

   Write the characteristic equation. For a linear recurrence with constant coefficients, guessing $a_n=r^n$ turns the relation into an algebraic equation. Substituting into $a_n-4a_{n-1}+4a_{n-2}=0$ and dividing by $r^{n-2}$:

   $r^2-4r+4=0\;\Longrightarrow\;(r-2)^2=0\;\Longrightarrow\;r=2\ \text{(double root)}$

2. 2

   Use the repeated-root form of the general solution. A double root gives only one geometric solution $2^n$, so the second independent solution picks up a factor of $n$:

   $a_n=(A+Bn)\,2^{n}$

   This is the step students most often get wrong — writing $A\cdot 2^n+B\cdot 2^n$ collapses to a single constant and cannot match two independent initial values.

3. 3

   Fit the two initial conditions. From $n=0$:

   $a_0=(A+0)\cdot 2^0=A=-2025$

   From $n=1$:

   $a_1=(A+B)\cdot 2=2026\;\Longrightarrow\;A+B=1013\;\Longrightarrow\;B=1013+2025=3038$

   So $a_n=(3038n-2025)\,2^{n}$. A quick sanity check: $a_2=(6076-2025)\cdot4=16204$, and directly $4a_1-4a_0=4(2026)-4(-2025)=16204$. They agree.

4. 4

   Evaluate at $n=2025$ and factor the coefficient.

   $a_{2025}=(3038\cdot 2025-2025)\,2^{2025}=2025\,(3038-1)\,2^{2025}=2025\cdot 3037\cdot 2^{2025}$

   Pulling out the common factor $2025$ before multiplying is what makes the last step clean — expanding first gives $6{,}149{,}925$, and you would then have to divide it back by $2025$.

5. 5

   Read off $b$ and $c$ and divide. Matching $a_{2025}=b\cdot 2^{c}$ gives $b=2025\cdot3037=6{,}149{,}925$ and $c=2025$, so

   $\frac{b}{c}=\frac{2025\cdot 3037}{2025}=3037$

   The factor $2025$ cancels exactly, which is the whole point of how the problem was built: the answer is $B-1=3037$.