Remainder....

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$\hspace{-16}\mathbb{I}$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}$$

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1057

Ketan Chandak ·May 25 '12 at 4:22

by some deduction we can prove a_n=32(3ⁿ-1)

now a₂₀₁₀=32(3²⁰¹⁰-1)

now 3⁵â‰¡1(mod 11)
so 3²⁰¹⁰â‰¡1(mod 11)
or
3²⁰¹¹â‰¡3(mod 11)

and thus the answer is 0....

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1708

man111 singh ·May 25 '12 at 4:26

Yes 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}$$

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Remainder....

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