1057by some deduction we can prove an=32(3n-1)
now a2010=32(32010-1)
now 35≡1(mod 11)
so 32010≡1(mod 11)
or
32011≡3(mod 11)
and thus the answer is 0....
\hspace{-16}\mathbb{I}$f $\bf{a_{0}=0}$ and $\bf{a_{n}=3\left(a_{n-1}+1\right)\forall n>1}$\\\\\\ Then Remainder when $\bf{a_{2010}}$ is Divided by $\bf{11}$
-
UP 0 DOWN 0 0 2
2 Answers
1057
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 $\bf{=0}$