Question

Let `a_(0)=0` and `a_(n)=3a_(n-1)+1` for `n ge 1`. Then the remainder obtained dividing `a_(2010)` by `11` is
A. `0`
B. `7`
C. `3`
D. `4`

Answer 1

Correct Answer - A
`(a)` `a_(n)=3a_(n-1)+1`
`a_(2010)=3a_(2009)+1`
`=3(3a_(2008)+1)+1=3^(2)a_(2008)+3+1`
`=3^(3)a_(2007)+3+3+1`
`=3^(4)a_(2006)+3+3+3+1`
`=3^(2010)a_(0)+(3+3+3+...2009"times")+1`
`=3xx2009+1`
`=6028`, which is divisible by `11`