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`
Text Solution
Verified by Experts
The correct Answer is:
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`
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CENGAGE-PROGRESSION AND SERIES-ARCHIVES (MATRIX MATCH TYPE )
