Question

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

• A. 483
• B. 528
• C. 575
• D. 624

Answer 1

Answer: (B)

528

Answer 2

an+2=3an+1−2an+1,∀ n≥0(a0=a1=0)
(an+2−an+1)−2(an+1−an)−1=0|
Put n = 0
(a2−a1)−2(a1−a0)−1=0
n = 1
(a3−a2)−2(a2−a1)−1=0
n = 2
(a4−a3)−2(a3−a2)−1=0
n=n
(an+2–an+1)−2(an+1−an)−1=0
On adding,
(an+2–a1)−2(aa+1−a0)−(n+1)=0
∴an+2−2an+1−(n+1)=0
n$\rightarrow$n–2
an–2an−1−n+1=0
Now,
a25a23−2a25a22−2a23a24+4a22a24
=a25(a23−2a22)−2a24(a23−2a22)
=(a25−2a24)(a23−2a22)
=24⋅22
=528
So, the correct option is (B): 528