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Let $a_{0}=0$ and $a_{x}=3 a_{n-1}+1$ for $n \geq 1$. Then the remainder obtained dividing aco ob by 11 is-
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$$
\begin{array}{l}
a_{n}=3 a_{n-1}+1 \\
a_{2010}=3 a_{2009}+1 \\
=3\left(3 a_{2008}+1\right)+1=3^{2} a_{2008}+3+1 \\
\quad=3^{3} a_{2007}+3+3+1
\end{array}
$$
$$
\begin{array}{l}
3^{2010} \mathrm{a}_{0}+\underbrace{(3+3+\ldots 3)}_{2009 \text { times }}+1 \\
=0+6027+1=6028
\end{array}
$$
Remainder $\left(\frac{6028}{11}\right)=0$
\begin{array}{l}
a_{n}=3 a_{n-1}+1 \\
a_{2010}=3 a_{2009}+1 \\
=3\left(3 a_{2008}+1\right)+1=3^{2} a_{2008}+3+1 \\
\quad=3^{3} a_{2007}+3+3+1
\end{array}
$$
$$
\begin{array}{l}
3^{2010} \mathrm{a}_{0}+\underbrace{(3+3+\ldots 3)}_{2009 \text { times }}+1 \\
=0+6027+1=6028
\end{array}
$$
Remainder $\left(\frac{6028}{11}\right)=0$
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