Given a1 - 12 a0 - Aa0 - a2 - 12 a1 - Aa1 and an - 1 - 12 an - Aan Find n- 2 where a - 0- A - 0- prove that an - -Aan - -A - a1 - -Aa1 - -A 2n - 1using mathematical induction-

We have prove #160; a_n √(A)a_n + √((A)) = (a_1 √(A)a_1 + √((A)) #160; )^2^n 1 #160; #160; #160;for all #160; nϵ ...

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Standard XI

Mathematics

Proof by mathematical induction

Given a1 = ...

Question

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.

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Solution

We have prove
$\frac{a_{n} - \sqrt{A}}{a_{n} + \sqrt{(A)}} = {(\frac{a_{1} - \sqrt{A}}{a_{1} + \sqrt{(A)}})}^{2^{n} - 1}$ for all $n ϵ N$
Clearly for n = 1, equality (1) holds .
Now from the given relation $a_{2} = \frac{1}{2} (a_{1} + \frac{A}{a_{1}})$
we have
$\frac{a_{2}}{\sqrt{(A)}} = \frac{1}{2 \sqrt{(A)}} (a_{1} + \frac{A}{a_{1}}) = \frac{a_{1}^{2} + A}{2 a_{1} \sqrt{(A)}}$
Using componendo and dividendo , this gives
$\frac{a_{2} - \sqrt{A}}{a_{2} + \sqrt{(A)}} = \frac{a_{1}^{2} + A - 2 a_{1} \sqrt{A}}{a_{1}^{2} + A + 2 a_{1} \sqrt{(A)}} = {(\frac{a_{1} - \sqrt{A}}{a_{1} + \sqrt{(A)}})}^{2}$
Hence the equality (1) holds for n = 1,
Now assume that (1) holds for n = m(m $\geq$ 2), that is, we assume
$\frac{a_{m} - \sqrt{A}}{a_{m} + \sqrt{(A)}} = {(\frac{a_{1} - \sqrt{A}}{a_{1} + \sqrt{(A)}})}^{2^{m} - 1}$ ........(2)
Then by the given relation
$a_{n + 1} = \frac{1}{2} (a_{n} + \frac{A}{a_{n}})$ we have
$\frac{a_{m + 1}}{\sqrt{(A)}} = \frac{1}{2 \sqrt{(A)}} (a_{m} + \frac{A}{a_{m}})$ (m $\geq$ 2) so that
$\frac{a_{m + 1 - \sqrt{A}}}{a_{m + 1} \sqrt{(A)}} = \frac{\frac{1}{2} (a_{m} + \frac{A}{a_{m}}) - \sqrt{A}}{\frac{1}{2} (a_{m} + \frac{A}{a_{m}}) + \sqrt{A}}$
$= \frac{a m^{2} - 2 a_{m} \sqrt{A} + A}{a m^{2} + 2 a_{m} \sqrt{A} + A}$
$= {(\frac{a_{m} - \sqrt{A}}{a_{m} + \sqrt{(A)}})}^{2} = {(\frac{a_{n} - \sqrt{A}}{a_{n} + \sqrt{(A)}})}^{2^{m}}$ by (2)
Hence the equality (1) holds for n = m + 1
Hence it holds for all $n ϵ N$

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Given a1=12(a0+Aa0),a2=12(a1+Aa1) and an+1=12(an+Aan) Find n≥2 where a > 0, A > 0, prove that an−√Aan+√(A)=(a1−√Aa1+√(A))2n−1using mathematical induction.