Let a_0=0 and a_n=3 a_{n-1}+1 for n geq 1. Th | vedclass

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(A) Given,$a_0=0$ and $a_n=3 a_{n-1}+1$ for $n \geq 1$.We can write the terms as:$a_1 = 3(0) + 1 = 1$$a_2 = 3(1) + 1 = 3 + 1$$a_3 = 3(3+1) + 1 = 3^2 +

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$0$

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(A) Given,$a_0=0$ and $a_n=3 a_{n-1}+1$ for $n \geq 1$.We can write the terms as:$a_1 = 3(0) + 1 = 1$$a_2 = 3(1) + 1 = 3 + 1$$a_3 = 3(3+1) + 1 = 3^2 + 3 + 1$In general,$a_n = 1 + 3 + 3^2 + \dots + 3^{n-1} = \frac{3^n - 1}{3 - 1} = \frac{3^n - 1}{2}$.We need to find the remainder of $a_{2010} = \frac{3^{2010} - 1}{2}$ when divided by $11$.This is equivalent to finding $x$ such that $\frac{3^{2010} - 1}{2} \equiv x \pmod{11}$.Multiplying by $2$,we have $3^{2010} - 1 \equiv 2x \pmod{11}$.By Fermat's Little Theorem,$3^{10} \equiv 1 \pmod{11}$.Since $2010 = 10 	imes 201$,we have $3^{2010} = (3^{10})^{201} \equiv 1^{201} \equiv 1 \pmod{11}$.Thus,$3^{2010} - 1 \equiv 1 - 1 \equiv 0 \pmod{11}$.Therefore,$2x \equiv 0 \pmod{11}$,which implies $x = 0$ since $2$ and $11$ are coprime.The remainder is $0$.

Let $a_0=0$ and $a_n=3 a_{n-1}+1$ for $n \geq 1$. Then,the remainder obtained by dividing $a_{2010}$ by $11$ is

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