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.

Given a1=12a0+Aa0, a2=12a1+Aa1 and an+1=12an+Aan for n≥2, where a>0,

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

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

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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.