Question

Let {an}ⁿ⁼⁰_∞ be a sequence such that a₀=a₁=0 and a_n+2=3a_n+1−2a_n+1,∀ n≥0. Then a₂₅a₂₃−2a₂₅a₂₂−2a₂₃a₂₄+4a₂₂a₂₄  is equal to

• A. 483
• B. 528
• C. 575
• D. 624

Answer 1

The Correct Option is B

The recurrence relation is a_n+2=3a_n+1−2a_n+1 for all n ≥ 0, with initial conditions a₀=a₁=0. This simplifies to a_n+2 = a_n+1 for n ≥ 0.
An alternative form of the equation is (a_n+2−a_n+1)−2(a_n+1−a_n)−1=0.
Substituting n = 0 yields (a₂−a₁)−2(a₁−a₀)−1=0.
Substituting n = 1 yields (a₃−a₂)−2(a₂−a₁)−1=0.
Substituting n = 2 yields (a₄−a₃)−2(a₃−a₂)−1=0.
For a general n, the equation is (a_n+2–a_n+1)−2(a_n+1−a_n)−1=0.
Upon summation, we get (a_n+2–a₁)−2(a_a+1−a₀)−(n+1)=0.
This simplifies to a_n+2−2a_n+1−(n+1)=0.
Replacing n with n–2, we have a_n–2a_n−1−n+1=0.
Consider the expression a₂₅a₂₃−2a₂₅a₂₂−2a₂₃a₂₄+4a₂₂a₂₄.
This can be factored as a₂₅(a₂₃−2a₂₂)−2a₂₄(a₂₃−2a₂₂).
Further factorization yields (a₂₅−2a₂₄)(a₂₃−2a₂₂).
The result of this expression is 24⋅22.
This equals 528.
Therefore, the correct option is (B): 528