Problem: Given a0=1,a1=3a_{0}=1, a_{1}=3a0=1,a1=3, and the general relation an2−an−1an+1=(−1)na_{n}^{2}-a_{n-1} a_{n+1}=(-1)^{n}an2−an−1an+1=(−1)n for n≥1n \geq 1n≥1. Then a3a_{3}a3 equals:
Answer Choices:
A. 13/2713 / 2713/27
B. 333333
C. 212121
D. 101010
E. −17-17−17 Solution:
When n=1a12−a0a2=(−1)1;a2=10n=1 \quad a_{1}^{2}-a_{0} a_{2}=(-1)^{1} ; a_{2}=10n=1a12−a0a2=(−1)1;a2=10
When n=2a22−a1a3=(−1)2;a3=33n=2 \quad a_{2}^{2}-a_{1} a_{3}=(-1)^{2} ; a_{3}=33n=2a22−a1a3=(−1)2;a3=33