A circle C1 is inscribed in a square S1 of side a1. Another square S2 of side a2 is incribed C1. A Circle C2 is inscribed in S1. The process is continued.... If a1=42, and s=a1+a2+.....∞ then (2-1)s=_______

Clearly BD=2a1⇒a1=BD2Clearly a2=B1D12….∴S=a1+a2+a3+……=42+4+22+2+…… =4211-12=4·2·22-1⇒(2-1)s=8

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A circle C₁ is inscribed in a square S₁ of side a₁. Another square S₂ of side a₂ is incribed C₁. A Circle C₂ is inscribed in S₁. The process is continued.... If a₁= $4 \sqrt{2}$ , and s=a₁+a₂+.....∞ then $(\sqrt{2} - 1) s =$ _______

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answer is 8.

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Detailed Solution

Clearly $BD = \sqrt{2} a_{1} \Rightarrow a_{1} = \frac{BD}{\sqrt{2}}$
Clearly $a_{2} = \frac{B_{1} D_{1}}{\sqrt{2}} \dots$ .
$∴ S = a_{1} + a_{2} + a_{3} + \dots \dots = 4 \sqrt{2} + 4 + 2 \sqrt{2} + 2 + \dots \dots$
$= 4 \sqrt{2} (\frac{1}{1 - \frac{1}{\sqrt{2}}}) = \frac{4 \cdot \sqrt{2} \cdot \sqrt{2}}{\sqrt{2} - 1}$