Question

A line segment initially 1 unit long grows according to the law
$1+\frac{\sqrt{2}}{4}+\frac{1}{4}+\frac{\sqrt{2}}{16}+\frac{1}{16}+\frac{\sqrt{2}}{64}+\frac{1}{64}+\ldots \ldots . . . \infty$
If the growth process continues indefinitely, then length of the line is equal to

• A. $\frac{4}{3}$
• B. $\frac{8}{3}$
• C. $\frac{4+\sqrt{2}}{3}$
• D. $\frac{2}{3}(4+\sqrt{2})$

Answer 1

Answer: (C)

$ S =\left(1+\frac{1}{4}+\frac{1}{16}+\frac{1}{64}+\ldots \ldots \infty\right)+\left(\frac{\sqrt{2}}{4}+\frac{\sqrt{2}}{16}+\frac{\sqrt{2}}{64}+\ldots \ldots \infty\right) $
$S =\frac{1}{1-\frac{1}{4}}+\frac{\frac{\sqrt{2}}{4}}{1-\frac{1}{4}}=\frac{4}{3}+\frac{4}{3}\left(\frac{\sqrt{2}}{4}\right)=\frac{4+\sqrt{2}}{3}$