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प्रश्न
\[\text{ Given } a_1 = \frac{1}{2}\left( a_0 + \frac{A}{a_0} \right), a_2 = \frac{1}{2}\left( a_1 + \frac{A}{a_1} \right) \text{ and } a_{n + 1} = \frac{1}{2}\left( a_n + \frac{A}{a_n} \right) \text{ for } n \geq 2, \text{ where } a > 0, A > 0 . \]
\[\text{ Prove that } \frac{a_n - \sqrt{A}}{a_n + \sqrt{A}} = \left( \frac{a_1 - \sqrt{A}}{a_1 + \sqrt{A}} \right) 2^{n - 1} .\]
उत्तर
\[\text{ Given: } a_1 = \frac{1}{2}\left( a_0 + \frac{A}{a_0} \right), a_2 = \frac{1}{2}\left( a_1 + \frac{A}{a_1} \right) \text{ and } a_{n + 1} = \frac{1}{2}\left( a_n + \frac{A}{a_n} \right) \text{ for } n \geq 2, \text { where } a > 0, A > 0 . \]
\[\text{ To prove: } \frac{a_n - \sqrt{A}}{a_n + \sqrt{A}} = \left( \frac{a_1 - \sqrt{A}}{a_1 + \sqrt{A}} \right)^{2^{n - 1}} \]
\[\text{ Proof: } \]
\[\text{ Let } p\left( n \right): \frac{a_n - \sqrt{A}}{a_n + \sqrt{A}} = \left( \frac{a_1 - \sqrt{A}}{a_1 + \sqrt{A}} \right)^{2^{n - 1}} \]
\[\text{ Step I: For } n = 1, \]
\[LHS = \frac{a_1 - \sqrt{A}}{a_1 + \sqrt{A}}\]
\[RHS = \left( \frac{a_1 - \sqrt{A}}{a_1 + \sqrt{A}} \right)^{2^{1 - 1}} = \frac{a_1 - \sqrt{A}}{a_1 + \sqrt{A}}\]
\[\text{ As, LHS = RHS } \]
\[\text{ So, it is true for n } = 1 . \]
\[\text{ Step II: For n } = k, \]
\[\text{ Let } p\left( k \right): \frac{a_k - \sqrt{A}}{a_k + \sqrt{A}} = \left( \frac{a_1 - \sqrt{A}}{a_1 + \sqrt{A}} \right)^{2^{k - 1}} \text{ be true for some values of k } \geq 2 . \]
\[\text{ Step III: For } n = k + 1, \]
\[p\left( k + 1 \right): \]
\[LHS = \frac{a_{k + 1} - \sqrt{A}}{a_{k + 1} + \sqrt{A}}\]
\[ = \frac{\frac{1}{2}\left( a_k + \frac{A}{a_k} \right) - \sqrt{A}}{\frac{1}{2}\left( a_k + \frac{A}{a_k} \right) + \sqrt{A}}\]
\[ = \frac{\frac{1}{2}\left( \frac{{a_k}^2 + A - 2 a_k \sqrt{A}}{a_k} \right)}{\frac{1}{2}\left( \frac{{a_k}^2 + A + 2 a_k \sqrt{A}}{a_k} \right)}\]
\[ = \frac{{a_k}^2 + A - 2 a_k \sqrt{A}}{{a_k}^2 + A + 2 a_k \sqrt{A}}\]
\[ = \frac{\left( a_k - \sqrt{A} \right)^2}{\left( a_k + \sqrt{A} \right)^2}\]
\[ = \left( \frac{a_k - \sqrt{A}}{a_k + \sqrt{A}} \right)^2 \]
\[ = \left[ \left( \frac{a_1 - \sqrt{A}}{a_1 + \sqrt{A}} \right)^{2^{k - 1}} \right]^2 \left( \text { Using step } II \right)\]
\[ = \left( \frac{a_1 - \sqrt{A}}{a_1 + \sqrt{A}} \right)^{2^{k - 1} \times 2} \]
\[ = \left( \frac{a_1 - \sqrt{A}}{a_1 + \sqrt{A}} \right)^{2^{k - 1 + 1}} \]
\[ = \left( \frac{a_1 - \sqrt{A}}{a_1 + \sqrt{A}} \right)^{2^k} \]
\[RHS = \left( \frac{a_1 - \sqrt{A}}{a_1 + \sqrt{A}} \right)^{2^{k + 1 - 1}} \]
\[ = \left( \frac{a_1 - \sqrt{A}}{a_1 + \sqrt{A}} \right)^{2^k} \]
\[As, LHS = RHS\]
\[\text{ So, it is also true for n } = k + 1 . \]
\[\text{ Hence, } \frac{a_n - \sqrt{A}}{a_n + \sqrt{A}} = \left( \frac{a_1 - \sqrt{A}}{a_1 + \sqrt{A}} \right)^{2^{n - 1}} .\]
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