Given a1-1-2a0-A-20-a2-1-2a1-A-a1 anda n-1-1-2an-A-an for n - 2- where a-0- A-0 - Prove that an-A-an-A-a1-A-a1-A2 n-1

A_1 = 1/2(a_0 + A/a_0), a_2 = 1/2(a_1 + A/a_1) and a_n+1 = 1/2(a_n + A/a_n) Let P(n) : a_n √(A)/a_n + √(A) = (a_1 √(A)/a_1 + √(A))^2n 1 For n = 1 a_1 √(A) ...

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Byju's Answer

Standard XII

Mathematics

Proof by mathematical induction

Given a1=1/2a...

Question

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

$a_{2} = \frac{1}{2} (a_{1} + \frac{A}{a_{1}}) a n d a_{n + 1} = \frac{1}{2} (a_{n} + \frac{A}{a_{n}})$ for n $\geq$ 2, where a > 0, A > 0. Prove that $\frac{a_{n} - \sqrt{A}}{a_{n} + \sqrt{A}} = {(\frac{a_{1} - \sqrt{A}}{a_{1} + \sqrt{A}})}^{2 n - 1}$

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Solution

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

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

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

For n = 1

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

$\frac{a_{1} - \sqrt{A}}{a_{1} + \sqrt{A}} = (\frac{a_{1} - \sqrt{A}}{a_{1} + \sqrt{A}})$

$\Rightarrow$ P(n) is true for n = 1

Let P(n) is true for n = k

$\frac{a_{k} - \sqrt{A}}{a_{k} + \sqrt{A}} = (\frac{a_{1} - \sqrt{A}}{a_{1} + \sqrt{A}})$ .........(i)

We have to show that

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

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

= ${⎡ ⎢ ⎢ ⎢ ⎣ \frac{\frac{\frac{1}{2} (a_{k} + \frac{A}{a_{k}} - \sqrt{A})}{1}}{2 (a_{k} + \frac{A}{a_{k}}) + \sqrt{A}} ⎤ ⎥ ⎥ ⎥ ⎦}^{2 n}$

$= {[\frac{(a_{k})^{2} + A - 2 a k \sqrt{A}}{(a_{k})^{2} + A - 2 a k \sqrt{A}}]}^{2 n}$

$= \frac{(a_{k} - \sqrt{A})^{2}}{(a_{k} + \sqrt{A})^{2}}$

$= {[\frac{a_{k} - \sqrt{A}}{a_{k} + \sqrt{A}}]}^{2^{1}}$

$= [\frac{a_{1} - \sqrt{A}}{a_{1} + \sqrt{A}}]^{2^{k}}$

$\Rightarrow$ P(n) is true for n = k + 1

$\Rightarrow$ P(n) is true for all $n ϵ N$ by PMI.

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Similar questions

Q. Given

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

and

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

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n \geq 2

where a > 0, A > 0, prove that

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

using mathematical induction.

Q. Given

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

for n ≥ 2, where a > 0, A > 0.
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\frac{a_{n} - \sqrt{A}}{a_{n} + \sqrt{A}} = (\frac{a_{1} - \sqrt{A}}{a_{1} + \sqrt{A}}) 2^{n - 1}

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Q. If

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\frac{a_{1} + a_{2 n}}{\sqrt{a_{1}} + \sqrt{a_{2}}} + \frac{a_{2} + a_{2 n - 1}}{\sqrt{a_{2}} + \sqrt{a_{3}}} + \frac{a_{3} + a_{2 n - 2}}{\sqrt{a_{3}} + \sqrt{a_{4}}} + . . . + \frac{a_{n} + a_{n + 1}}{\sqrt{a_{n}} + \sqrt{a_{n + 1}}} =

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Given a1=12(a0+Aa0), a2=12(a1+Aa1)andan+1=12(an+Aan) for n ≥ 2, where a > 0, A > 0. Prove that an−√Aan+√A=(a1−√Aa1+√A)2n−1