What is the sum of the infinite terms of the | vedclass

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Question

(A) The given geometric progression is $a, ar, ar^2, \dots$ where $a = \frac{\sqrt{2} + 1}{\sqrt{2} - 1}$.Rationalizing $a$: $a = \frac{(\sqrt{2} + 1)

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$\sqrt{2}(\sqrt{2} + 1)^2$

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(A) The given geometric progression is $a, ar, ar^2, \dots$ where $a = \frac{\sqrt{2} + 1}{\sqrt{2} - 1}$.Rationalizing $a$: $a = \frac{(\sqrt{2} + 1)^2}{2 - 1} = (\sqrt{2} + 1)^2 = 3 + 2\sqrt{2}$.The second term is $\frac{1}{2 - \sqrt{2}} = \frac{1}{\sqrt{2}(\sqrt{2} - 1)} = \frac{\sqrt{2} + 1}{\sqrt{2}(2 - 1)} = \frac{\sqrt{2} + 1}{\sqrt{2}} = 1 + \frac{1}{\sqrt{2}}$.The common ratio $r = \frac{T_2}{T_1} = \frac{1}{\sqrt{2}(\sqrt{2} + 1)} 	imes \frac{\sqrt{2} - 1}{\sqrt{2} + 1} = \frac{1}{\sqrt{2}(\sqrt{2} + 1)} 	imes \frac{\sqrt{2} - 1}{1} = \frac{\sqrt{2} - 1}{\sqrt{2}(\sqrt{2} + 1)} = \frac{(\sqrt{2} - 1)^2}{\sqrt{2}(2 - 1)} = \frac{3 - 2\sqrt{2}}{\sqrt{2}} = \frac{3}{\sqrt{2}} - 2$.Alternatively,$r = \frac{1}{\sqrt{2}(\sqrt{2} + 1)} = \frac{\sqrt{2} - 1}{\sqrt{2}} = 1 - \frac{1}{\sqrt{2}}$.The sum of infinite terms $S = \frac{a}{1 - r} = \frac{(\sqrt{2} + 1)^2}{1 - (1 - \frac{1}{\sqrt{2}})} = \frac{(\sqrt{2} + 1)^2}{\frac{1}{\sqrt{2}}} = \sqrt{2}(\sqrt{2} + 1)^2$.

What is the sum of the infinite terms of the geometric progression $\frac{\sqrt{2} + 1}{\sqrt{2} - 1}, \frac{1}{2 - \sqrt{2}}, \frac{1}{2}, \dots$?

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