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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 - |
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Answer» Correct Answer - 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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