1708Yes Ketan You are Right
My solution..........
\hspace{-16}$Here $\bf{a_{n}=3\left(a_{n-1}+1\right)}$\\\\\\ $\bf{\left(a_{n}+\frac{1}{2}\right)=3.\left(a_{n-1}+\frac{1}{2}\right)}$\\\\\\ Using Recursively, We Get\\\\\\ $\bf{\left(a_{n}+\frac{1}{2}\right)=3^n.\left(a_{0}+\frac{1}{2}\right)=\frac{3^n}{2}}$\\\\\\ So $\bf{a_{n}=\frac{3^n-1}{2}}$, bcz $\bf{a_{0}=0}$ (Given)\\\\\\ So $\bf{a_{2010}=\frac{3^{2010}-1}{2}=\frac{(3^{10})^{201}-1}{2}=0(mod (11))}$\\\\\\ So Remainder is $\bf{=0}$