Given a 1-12 a 0-A a 0- a 2-12 a 1-A a 1 and a n-1-12 a n- Aan for n - 2- where a-0- A-0 Prove that an-Aan - A - a 1- Aa 1- A 2 n -1

Let:Pn:an Aan+A=a1 Aa1+A2n 1Step #160;I #160; #160;P(1):a1 Aa1+A=a1 Aa1+A21 1 #160; #160; #160; #160; #160; #160; #160; #160; #160;(whic ...

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Given a 1=12 ...

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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.
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

$Let : P (n) : \frac{a_{n} - \sqrt{A}}{a_{n} + \sqrt{A}} = {(\frac{a_{1} - \sqrt{A}}{a_{1} + \sqrt{A}})}^{2^{n - 1}} Step I P (1) : (\frac{a_{1} - \sqrt{A}}{a_{1} + \sqrt{A}}) = {(\frac{a_{1} - \sqrt{A}}{a_{1} + \sqrt{A}})}^{2^{1 - 1}} (which is true) P (2) : (\frac{a_{2} - \sqrt{A}}{a_{2} + \sqrt{A}}) = (\frac{\frac{1}{2} (a_{1} + \frac{A}{a_{1}}) - \sqrt{A}}{\frac{1}{2} (a_{1} + \frac{A}{a_{1}}) + \sqrt{A}}) = (\frac{a_{1} + \frac{A}{a_{1}} - 2 \sqrt{A}}{a_{1} + \frac{A}{a_{1}} + 2 \sqrt{A}}) = (\frac{a_{1} + A - 2 \sqrt{A}}{a_{1} + A + 2 \sqrt{A}}) = {(\frac{a_{1} - \sqrt{A}}{a_{1} + \sqrt{A}})}^{2^{2 - 1}} Thus, P (1) and P (2) are true . Step II Let P (k) be true . Now, \frac{a_{k} - \sqrt{A}}{a_{k} + \sqrt{A}} = {(\frac{a_{1} - \sqrt{A}}{a_{1} + \sqrt{A}})}^{2^{k - 1}} . . . . . (i) and P (k + 1) : \frac{a_{k + 1} - \sqrt{A}}{a_{k + 1} + \sqrt{A}} = \frac{\frac{1}{2} (a_{k} + \frac{A}{a_{k}}) - \sqrt{A}}{\frac{1}{2} (a_{k} + \frac{A}{a_{k}}) + \sqrt{A}} = \frac{(a_{k} + \frac{A}{a_{k}}) - 2 \sqrt{A}}{(a_{k} + \frac{A}{a_{k}}) + 2 \sqrt{A}} = \frac{{(\sqrt{a_{k}} - \sqrt{\frac{A}{a_{k}}})}^{2}}{{(\sqrt{a_{k}} + \sqrt{\frac{A}{a_{k}}})}^{2}} = {(\frac{\sqrt{a_{k}} - \sqrt{\frac{A}{a_{k}}}}{\sqrt{a_{k}} + \sqrt{\frac{A}{a_{k}}}})}^{2} = {(\frac{a_{k} - \sqrt{A}}{a_{k} + \sqrt{A}})}^{2} = {[{(\frac{a_{1} - \sqrt{A}}{a_{1} + \sqrt{A}})}^{2^{k - 1}}]}^{2} = {(\frac{a_{1} - \sqrt{A}}{a_{1} + \sqrt{A}})}^{2^{k}} = {(\frac{a_{1} - \sqrt{A}}{a_{1} + \sqrt{A}})}^{2^{(k + 1) - 1}} Thus, P (k + 1) is also true .$

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

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Similar questions

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

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}})

Find

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. If

a_{r} > 0; \forall r, n \in N

and

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

are in A.P, then

\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}}} =

Given that a_i $>$ 0 and i belongs to a set of natural numbers. If $a_{1}, a_{2}, a_{3} . . . . . a_{2 n}$ are in AP, then find the value of $\frac{a_{1} + a_{2 n}}{\sqrt{a_{1}} + \sqrt{a_{2}}} + \frac{a_{2} + a_{2 n} - 1}{{\sqrt{a}}_{2} + \sqrt{a_{3}}} . . . . . . . . . . . . + \frac{a_{n} + a_{n + 1}}{\sqrt{a_{n}} + \sqrt{a_{n + 1}}}$

Q. If

a_{r} > 0, r \in N

and

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

are in A.P. then

\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}}}

is equal to

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Given a1=12a0+Aa0,a2=12a1+Aa1 and an+1=12an+Aan for n ≥ 2, where a > 0, A > 0. Prove that an-Aan+A=a1-Aa1+A2n-1