Let a(0)=0 and a(n)=3a(n-1)+1 for n ge 1...

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

`0`

`7`

`3`

`4`

Text Solution

Verified by Experts

The correct Answer is:

`(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`

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