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

Given: #160;a1=12a0+Aa0, #160;a2=12a1+Aa1 #160;and #160;an+1=12an+Aan #160;for #160;n #8805;2, #160;where #160;a #62;0, #160;A #62;0.To ...

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Given a 1=12 ...

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

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$Given : a_{1} = \frac{1}{2} (a_{0} + \frac{A}{a_{0}}), a_{2} = \frac{1}{2} (a_{1} + \frac{A}{a_{1}}) and a_{n + 1} = \frac{1}{2} (a_{n} + \frac{A}{a_{n}}) for n \geq 2, where a > 0, A > 0 . To prove : \frac{a_{n} - \sqrt{A}}{a_{n} + \sqrt{A}} = {(\frac{a_{1} - \sqrt{A}}{a_{1} + \sqrt{A}})}^{2^{n - 1}} Proof : Let p (n) : \frac{a_{n} - \sqrt{A}}{a_{n} + \sqrt{A}} = {(\frac{a_{1} - \sqrt{A}}{a_{1} + \sqrt{A}})}^{2^{n - 1}} Step I : For n = 1, LHS = \frac{a_{1} - \sqrt{A}}{a_{1} + \sqrt{A}} R H S = {(\frac{a_{1} - \sqrt{A}}{a_{1} + \sqrt{A}})}^{2^{1 - 1}} = \frac{a_{1} - \sqrt{A}}{a_{1} + \sqrt{A}} As, LHS = RHS So, it is true for n = 1 . Step II : For n = k, Let p (k) : \frac{a_{k} - \sqrt{A}}{a_{k} + \sqrt{A}} = {(\frac{a_{1} - \sqrt{A}}{a_{1} + \sqrt{A}})}^{2^{k - 1}} be true for some values of k \geq 2 . Step III : For n = k + 1, p (k + 1) : LHS = \frac{a_{k + 1} - \sqrt{A}}{a_{k + 1} + \sqrt{A}} = \frac{\frac{1}{2} (a_{k} + \frac{A}{a_{k}}) - \sqrt{A}}{\frac{1}{2} (a_{k} + \frac{A}{a_{k}}) + \sqrt{A}} = \frac{\frac{1}{2} (\frac{{a_{k}}^{2} + A - 2 a_{k} \sqrt{A}}{a_{k}})}{\frac{1}{2} (\frac{{a_{k}}^{2} + A + 2 a_{k} \sqrt{A}}{a_{k}})} = \frac{{a_{k}}^{2} + A - 2 a_{k} \sqrt{A}}{{a_{k}}^{2} + A + 2 a_{k} \sqrt{A}} = \frac{{(a_{k} - \sqrt{A})}^{2}}{{(a_{k} + \sqrt{A})}^{2}} = {(\frac{a_{k} - \sqrt{A}}{a_{k} + \sqrt{A}})}^{2} = {[{(\frac{a_{1} - \sqrt{A}}{a_{1} + \sqrt{A}})}^{2^{k - 1}}]}^{2} (Using step II) = {(\frac{a_{1} - \sqrt{A}}{a_{1} + \sqrt{A}})}^{2^{k - 1} \times 2} = {(\frac{a_{1} - \sqrt{A}}{a_{1} + \sqrt{A}})}^{2^{k - 1 + 1}} = {(\frac{a_{1} - \sqrt{A}}{a_{1} + \sqrt{A}})}^{2^{k}} RHS = {(\frac{a_{1} - \sqrt{A}}{a_{1} + \sqrt{A}})}^{2^{k + 1 - 1}} = {(\frac{a_{1} - \sqrt{A}}{a_{1} + \sqrt{A}})}^{2^{k}} As, LHS = RHS So, it is also true for n = k + 1 . Hence, \frac{a_{n} - \sqrt{A}}{a_{n} + \sqrt{A}} = {(\frac{a_{1} - \sqrt{A}}{a_{1} + \sqrt{A}})}^{2^{n - 1}} .$

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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.

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; \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}}} =

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.