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

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:
C
`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^(2010)a_(0) + ubrace((3+3+"....."3))_("2009 times")+1`
` = 0 + 6027 + 1 = 6028`
Remainder `((6028)/(11)) = 0`
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