Given a_1=1 2 (a_0+ A a_0 ),a_2=1 2 (a_1+ A a_1 ) and a_ n+1 =1 2 (a_n+ A a_n ) for n 2, Where a > 0, A > 0. Prove that a_ n - A a_ n + A = ( a_ 1 - A a_ 1 + A )^ 2^ n-1

a_1=1 2 (a_0+ A a_0 ),a_2=1 2 (a_1+ A a_1 ) and a_ n+1 =1 2 (a_n+ A a_n ) Let P(n): a_ n - A a_ n + A = ( a_ 1 - A a_ 1 + A )^ 2^ n-1 For n = 1 a_ 1 - A a_ 1 + A = ( a_ 1 - A a_ 1 + A )^ 2^ n-1 a_ 1 - A a_ 1 + A = ( a_ 1 - A a_ 1 + A ) ⇒ p(n) is true for n = 1 Let p(n) is true for n = k, a_ k - A a_ k + A = ( a_ 1 - A a_ 1 + A ) ...(1) We have to show that, a_ k+1 - A a_ k+1 + A = ( a_ 1 - A a_ 1 + A )^ 2k ( a_ k+1 - A a_ k+1 + A )^ 2^0 = 1 2 (a_k+ A a_k )- A 1 2 (a_k+ A a_k )+ A ^ 2^0 = [ (a_k)^2+A-2a_k A (a_k)^2+A+2a_k A ]^ 2^0 = (a_ k - A )^2 ( a_ k + A )^2 = [ a_ k - A a_ k + A ]^ 2^1 = [ a_ 1 - A a_ 1 + A ]^ 2^k ⇒ P(n) is true for n = k + 1 ⇒ P(n) is true for n by PM

Given a_1=1 2 (a_0+ A a_0 ),a_2=1 2 (a_1+ A a_1 ) and a_ n+1 =1 2…

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Question

Given $\text{a}_1=\frac{1}{2}\Big(\text{a}_0+\frac{\text{A}}{\text{a}_0}\Big),\text{a}_2=\frac{1}{2}\Big(\text{a}_1+\frac{\text{A}}{\text{a}_1}\Big)$ and $\text{a}_{\text{n}+1}=\frac{1}{2}\Big(\text{a}_\text{n}+\frac{\text{A}}{\text{a}_\text{n}}\Big)$ for $\text{n}\geq2,$ Where a > 0, A > 0. Prove that $\frac{\text{a}_{\text{n}}-\sqrt{\text{A}}}{{\text{a}_{\text{n}}+\sqrt{\text{A}}}}=\Bigg(\frac{\text{a}_{\text{1}}-\sqrt{\text{A}}}{\text{a}_{\text{1}}+\sqrt{\text{A}}}\Bigg)^{2^{\text{n}-1}}$

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Answer

$\text{a}_1=\frac{1}{2}\Big(\text{a}_0+\frac{\text{A}}{\text{a}_0}\Big),\text{a}_2=\frac{1}{2}\Big(\text{a}_1+\frac{\text{A}}{\text{a}_1}\Big)$ and $\text{a}_{\text{n}+1}=\frac{1}{2}\Big(\text{a}_\text{n}+\frac{\text{A}}{\text{a}_\text{n}}\Big)$
Let P(n): $\frac{\text{a}_{\text{n}}-\sqrt{\text{A}}}{{\text{a}_{\text{n}}+\sqrt{\text{A}}}}=\Bigg(\frac{\text{a}_{\text{1}}-\sqrt{\text{A}}}{\text{a}_{\text{1}}+\sqrt{\text{A}}}\Bigg)^{2^{\text{n}-1}}$
For n = 1
$\frac{\text{a}_{\text{1}}-\sqrt{\text{A}}}{{\text{a}_{\text{1}}+\sqrt{\text{A}}}}=\Bigg(\frac{\text{a}_{\text{1}}-\sqrt{\text{A}}}{\text{a}_{\text{1}}+\sqrt{\text{A}}}\Bigg)^{2^{\text{n}-1}}$
$\frac{\text{a}_{\text{1}}-\sqrt{\text{A}}}{{\text{a}_{\text{1}}+\sqrt{\text{A}}}}=\Bigg(\frac{\text{a}_{\text{1}}-\sqrt{\text{A}}}{\text{a}_{\text{1}}+\sqrt{\text{A}}}\Bigg)$
⇒ p(n) is true for n = 1
Let p(n) is true for n = k,
$\frac{\text{a}_{\text{k}}-\sqrt{\text{A}}}{{\text{a}_{\text{k}}+\sqrt{\text{A}}}}=\Bigg(\frac{\text{a}_{\text{1}}-\sqrt{\text{A}}}{\text{a}_{\text{1}}+\sqrt{\text{A}}}\Bigg) \ ...(1)$
We have to show that,
$\frac{\text{a}_{\text{k}+1}-\sqrt{\text{A}}}{{\text{a}_{\text{k}+1}+\sqrt{\text{A}}}}=\Bigg(\frac{\text{a}_{\text{1}}-\sqrt{\text{A}}}{\text{a}_{\text{1}}+\sqrt{\text{A}}}\Bigg)^{2\text{k}}$
$\Bigg(\frac{\text{a}_{\text{k}+1}-\sqrt{\text{A}}}{{\text{a}_{\text{k}+1}+\sqrt{\text{A}}}}\Bigg)^{2^0}$
$=\begin{bmatrix}\frac{\frac{\frac{1}{2}\Big(\text{a}_\text{k}+\frac{\text{A}}{\text{a}_\text{k}}\Big)-\sqrt{\text{A}}}{1}}{2\Big(\text{a}_\text{k}+\frac{\text{A}}{\text{a}_\text{k}}\Big)+\sqrt{\text{A}}} \end{bmatrix}^{2^0}$
$=\Bigg[\frac{(\text{a}_\text{k})^2+\text{A}-2\text{a}_\text{k}\sqrt{\text{A}}}{(\text{a}_\text{k})^2+\text{A}+2\text{a}_\text{k}\sqrt{\text{A}}}\Bigg]^{2^0}$
$=\frac{\Big(\text{a}_{\text{k}}-\sqrt{\text{A}}\Big)^2}{\Big({\text{a}_{\text{k}}+\sqrt{\text{A}}\Big)^2}}$
$=\Bigg[\frac{\text{a}_{\text{k}}-\sqrt{\text{A}}}{{\text{a}_{\text{k}}+\sqrt{\text{A}}}}\Bigg]^{2^1}$
$=\Bigg[\frac{\text{a}_{\text{1}}-\sqrt{\text{A}}}{{\text{a}_{\text{1}}+\sqrt{\text{A}}}}\Bigg]^{2^\text{k}}$
⇒ P(n) is true for n = k + 1
⇒ P(n) is true for $\text{n}\in\text{N}$ by PM

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