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A.G.P

Let an be a s...

Question

$Let {a_{n}} be a sequence such that a_{0} = 1, a_{1} = 0, a_{n} = 3 a_{n - 1} - 2 a_{n - 2} . Then, which of the following is a correct statement?$

a_{45} = 2^{45}

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a_{51} = 2^{51} - 2

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a_{48} = 2 (1 - 2^{47})

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a_{49} = \sqrt{2} - \sqrt{2^{49}}

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Solution

The correct option is C $a_{48} = 2 (1 - 2^{47})$
$a_{0} = 1, a_{1} = 0 and a_{n} = 3 a_{n - 1} - 2 a_{n - 2} ∴ (a_{n} - 2 a_{n - 1}) = (a_{n - 1} - 2 a_{n - 2}) = k (let) So, a_{1} - 2 a_{0} = k = - 2 ∴ a_{n} = 2 a_{n - 1} - 2 \Rightarrow a_{n} - 2 = 2 (a_{n - 2} - 2) Let b_{n} = 2 \cdot b_{n - 1} ∴ b_{0}, b_{1}, b_{2}, \dots, b_{n} are in G.P. . b_{n} = b_{0} \cdot 2^{n} \Rightarrow b_{n} = (- 1)^{n} \cdot 2^{n} \Rightarrow a_{n} - 2 = - 2^{n} ∴ a_{n} = 2 - 2^{n} ∴ a_{48} = 2 (1 - 2^{47})$

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Similar questions

Let {a_{n}} be a sequence such that a_{0} = 1, a_{1} = 0, a_{n} = 3 a_{n - 1} - 2 a_{n - 2} . Then, which of the following is a correct statement?

a_{1} = 1

a_{n + 1} = \frac{4 + 3 a_{n}}{3 + 2 a_{n}}, n \geq 1

lim n \to \infty a_{n} = n

a_{1} = \frac{1}{2} (a_{0} + \frac{A}{a_{0}}), a_{2} = \frac{1}{2} (a_{1} + \frac{A}{a_{1}})

a_{n + 1} = \frac{1}{2} (a_{n} + \frac{A}{a_{n}})

n \geq 2

\frac{a_{n} - \sqrt{A}}{a_{n} + \sqrt{(A)}} = {(\frac{a_{1} - \sqrt{A}}{a_{1} + \sqrt{(A)}})}^{2^{n - 1}}

a_{0} = 0

a_{n} = 3 a_{n - 1} + 1

n \geq 1

a_{2010}

a_{k} > 0 \forall k \geq 1 a n d a_{1}, a_{2}, a_{3} . . .

Δ_{n} = 0 \forall n \geq 1

a_{1}, a_{2}, a_{3} . . .

Δ_{n} = 0 \forall n \geq 1

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