The recurrence relation is an+2=3an+1−2an+1 for all n ≥ 0, with initial conditions a0=a1=0. This simplifies to an+2 = an+1 for n ≥ 0. An alternative form of the equation is (an+2−an+1)−2(an+1−an)−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 (an+2–an+1)−2(an+1−an)−1=0. Upon summation, we get (an+2–a1)−2(aa+1−a0)−(n+1)=0. This simplifies to an+2−2an+1−(n+1)=0. Replacing n with n–2, we have an–2an−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
