Let a 1-2- a n -1- a n 2- a n -1 for n - 1- then - n -1-1- a n

We have a_n+1=a_n^2 a_n+1 =(a_n 1/2)^2+3/4 gt;0 So a_n ≠ 0 for n ∈ N ∴ a_n is an increasing function. a_n+1 1=a_n^2 a_n ⇒1/a_n+1 1=1/a_n^2 a_n ⇒a_n/a_n+1 ...

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Standard XII

Mathematics

Method of Difference

Let a 1=2, a ...

Question

$Let a_{1} = 2, a_{n + 1} = a_{n}^{2} - a_{n} + 1 for n \geq 1, then \infty \sum n = 1 \frac{1}{a_{n}} is$

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Solution

$We have a_{n + 1} = a_{n}^{2} - a_{n} + 1 = {(a_{n} - \frac{1}{2})}_{n}^{2} + \frac{3}{4} > 0 So a_{n} \neq 0 for n \in N ∴ a_{n} is an increasing function. a_{n + 1} - 1 = a_{n}^{2} - a_{n} \Rightarrow \frac{1}{a_{n + 1} - 1} = \frac{1}{a_{n}^{2} - a_{n}} \Rightarrow \frac{a_{n}}{a_{n + 1} - 1} = \frac{1}{a_{n} - 1} \Rightarrow \frac{a_{n}}{a_{n + 1} - 1} - \frac{1}{a_{n}} = \frac{1}{a_{n} - 1} - \frac{1}{a_{n}} \Rightarrow \frac{a_{n}^{2} - a_{n + 1} + 1}{a_{n} (a_{n + 1} - 1)} = \frac{1}{a_{n} - 1} - \frac{1}{a_{n}} \Rightarrow \frac{1}{(a_{n + 1} - 1)} = \frac{1}{a_{n} - 1} - \frac{1}{a_{n}} ∴ \infty \sum n = 1 \frac{1}{a_{n}} = \infty \sum n = 1 (\frac{1}{a_{n} - 1} - \frac{1}{a_{n + 1} - 1}) \Rightarrow \infty \sum n = 1 \frac{1}{a_{n}} = (\frac{1}{a_{1} - 1}) + (lim n \to \infty \frac{1}{a_{n + 1} - 1}) \Rightarrow \infty \sum n = 1 \frac{1}{a_{n}} = 1$

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Similar questions

Q. Consider the sequence

a_{n}

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a_{1} = \frac{1}{2}, a_{n + 1} = a_{n}^{2} + a_{n}

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S_{n} = \frac{1}{a_{1} + 1} + \frac{1}{a_{2} + 1} + . . . + \frac{1}{a_{n} + 1}

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(iii) a₁ = a₂ = 2, a_n = a_n_{− 1} − 1, n > 2

Q. Let a sequence be defined by

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a_{n} = a_{n - 1} + a_{n - 2}

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Let a1=2, an+1=a2n−an+1 for n≥1, then ∞∑n=11an is

$Let a_{1} = 2, a_{n + 1} = a_{n}^{2} - a_{n} + 1 for n \geq 1, then \infty \sum n = 1 \frac{1}{a_{n}} is$