A. 399-200
B. 399+200
C. 3100-200
D. 3100+200
a1 = 1.
an + 1 = 3an - 2 + 4n
a2 = 3 – 2 + 4 = 5
a3 = 15 – 2 + 8 = 21
a4 = 63 – 2 + 12 = 73.
Now, our options involve 3n followed by some addition or subtraction.
So a2 = 5, can be written as 31 + 2 and 32 - 4.
a3 = 21 can be written as 32 + 12 and 33 – 6.
a4 = 63 can be written as 33 + 36 and 34 – 8.
We see that the subtraction trend is repeating, as 3n – 2*n. So, a100 is 3100 – 200.