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प्रश्न
Given \[a_1 = \frac{1}{2}\left( a_0 + \frac{A}{a_0} \right), a_2 = \frac{1}{2}\left( a_1 + \frac{A}{a_1} \right) \text{ and } a_{n + 1} = \frac{1}{2}\left( a_n + \frac{A}{a_n} \right)\] for n ≥ 2, where a > 0, A > 0.
Prove that \[\frac{a_n - \sqrt{A}}{a_n + \sqrt{A}} = \left( \frac{a_1 - \sqrt{A}}{a_1 + \sqrt{A}} \right) 2^{n - 1}\]
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उत्तर
\[\text{ Let } : \]
\[P\left( n \right): \frac{a_n - \sqrt{A}}{a_n + \sqrt{A}} = \left( \frac{a_1 - \sqrt{A}}{a_1 + \sqrt{A}} \right)^{2^{n - 1}} \]
\[\text{ Step } I \]
\[P(1): \left( \frac{a_1 - \sqrt{A}}{a_1 + \sqrt{A}} \right) = \left( \frac{a_1 - \sqrt{A}}{a_1 + \sqrt{A}} \right)^{2^{1 - 1}} (\text{ which is true } )\]
\[P(2): \left( \frac{a_2 - \sqrt{A}}{a_2 + \sqrt{A}} \right) = \left( \frac{\frac{1}{2}\left( a_1 + \frac{A}{a_1} \right) - \sqrt{A}}{\frac{1}{2}\left( a_1 + \frac{A}{a_1} \right) + \sqrt{A}} \right) = \left( \frac{a_1 + \frac{A}{a_1} - 2\sqrt{A}}{a_1 + \frac{A}{a_1} + 2\sqrt{A}} \right) = \left( \frac{a_1 + A - 2\sqrt{A}}{a_1 + A + 2\sqrt{A}} \right) = \left( \frac{a_1 - \sqrt{A}}{a_1 + \sqrt{A}} \right)^{2^{2 - 1}} \]
\[\text{ Thus, P(1) and P(2) are true } . \]
\[\text{ Step } II \]
\[\text{ Let P(k) be true } . \]
\[\text{ Now, } \]
\[\frac{a_k - \sqrt{A}}{a_k + \sqrt{A}} = \left( \frac{a_1 - \sqrt{A}}{a_1 + \sqrt{A}} \right)^{2^{k - 1}} . . . . . (i)\]
\[\text{ and } \]
\[P(k + 1): \frac{a_{k + 1} - \sqrt{A}}{a_{k + 1} + \sqrt{A}} = \frac{\frac{1}{2}\left( a_k + \frac{A}{a_k} \right) - \sqrt{A}}{\frac{1}{2}\left( a_k + \frac{A}{a_k} \right) + \sqrt{A}}\]
\[ = \frac{\left( a_k + \frac{A}{a_k} \right) - 2\sqrt{A}}{\left( a_k + \frac{A}{a_k} \right) + 2\sqrt{A}}\]
\[ = \frac{\left( \sqrt{a_k} - \sqrt{\frac{A}{a_k}} \right)^2}{\left( \sqrt{a_k} + \sqrt{\frac{A}{a_k}} \right)^2}\]
\[ = \left( \frac{\sqrt{a_k} - \sqrt{\frac{A}{a_k}}}{\sqrt{a_k} + \sqrt{\frac{A}{a_k}}} \right)^2 \]
\[ = \left( \frac{a_k - \sqrt{A}}{a_k + \sqrt{A}} \right)^2 \]
\[ = \left[ \left( \frac{a_1 - \sqrt{A}}{a_1 + \sqrt{A}} \right)^{2^{k - 1}} \right]^2 \]
\[ = \left( \frac{a_1 - \sqrt{A}}{a_1 + \sqrt{A}} \right)^{2^k} \]
\[ = \left( \frac{a_1 - \sqrt{A}}{a_1 + \sqrt{A}} \right)^{2^\left( k + 1 \right) - 1} \]
\[\text{ Thus, P(k + 1) is also true .} \]
\[= \frac{a_k \left( a_k + \sqrt{A} \right) \left( \frac{a_k - \sqrt{A}}{a_k + \sqrt{A}} \right)^{2^{k - 1}} - \sqrt{A}\left( a_k - \sqrt{A} \right)}{\left( a_k + \sqrt{A} \right)^2}\]
\[ = \frac{a_k \left( a_k + \sqrt{A} \right) \left( \frac{a_k - \sqrt{A}}{a_k + \sqrt{A}} \right)^{2^{k - 1}} - \sqrt{A}\left( a_k + \sqrt{A} \right) \left( \frac{a_k - \sqrt{A}}{a_k + \sqrt{A}} \right)^{2^{k - 1}}}{\left( a_k + \sqrt{A} \right)^2} \left[ \text{ Using } (i) \right]\]
\[ = \frac{\left( a_k + \sqrt{A} \right) \left( \frac{a_k - \sqrt{A}}{a_k + \sqrt{A}} \right)^{2^{k - 1}} \left( a_k - \sqrt{A} \right)}{\left( a_k + \sqrt{A} \right)^2}\]
\[ = \frac{\left( \frac{a_k - \sqrt{A}}{a_k + \sqrt{A}} \right)^{2^{k - 1}} \left( a_k - \sqrt{A} \right)}{\left( a_k + \sqrt{A} \right)}\]
\[ = \left( \frac{a_k - \sqrt{A}}{a_k + \sqrt{A}} \right)^{2^{k - 1}} \left( \frac{a_k - \sqrt{A}}{a_k + \sqrt{A}} \right)\]
\[ = \left( \frac{a_k - \sqrt{A}}{a_k + \sqrt{A}} \right)^\left( 2^{k - 1} + 1 \right) \]
\[ = \left( \left( \frac{a_1 - \sqrt{A}}{a_1 + \sqrt{A}} \right)^{2^{k - 1}} \right)^\left( 2^{k - 1} + 1 \right) \]
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