# 1952 AHSME Problems/Problem 50

## Problem

A line initially 1 inch long grows according to the following law, where the first term is the initial length. $$1+\frac{1}{4}\sqrt{2}+\frac{1}{4}+\frac{1}{16}\sqrt{2}+\frac{1}{16}+\frac{1}{64}\sqrt{2}+\frac{1}{64}+\cdots$$

If the growth process continues forever, the limit of the length of the line is: $\textbf{(A) } \infty\qquad \textbf{(B) } \frac{4}{3}\qquad \textbf{(C) } \frac{8}{3}\qquad \textbf{(D) } \frac{1}{3}(4+\sqrt{2})\qquad \textbf{(E) } \frac{2}{3}(4+\sqrt{2})$

## Solution

We can rewrite our sum as the sum of two infinite geometric sequences. $$1 + \frac{1}{4}\sqrt{2} + \frac{1}{4} + \frac{1}{16}\sqrt{2} + \frac{1}{16} + ... =$$ $$(1 + \frac{1}{4} + \frac{1}{16} + \frac{1}{64} + ...) + (\frac{1}{4}\sqrt{2} + \frac{1}{16}\sqrt{2} + \frac{1}{64}\sqrt{2} + ...)$$ $$(1 + \frac{1}{4} + \frac{1}{16} + \frac{1}{64} + ...) + \sqrt{2}(\frac{1}{4} + \frac{1}{16} + \frac{1}{64} + ...)$$ We now take the sum of each of the infinite geometric sequences separately $$(1 + \frac{1}{4} + \frac{1}{16} + \frac{1}{64} + ...) + \sqrt{2}(\frac{1}{4} + \frac{1}{16} + \frac{1}{64} + ...) =$$ $$\frac{1}{1 - \frac{1}{4}} + \sqrt{2}(\frac{\frac{1}{4}}{1 - \frac{1}{4}})$$ $$\frac{4}{3} + \frac{\sqrt{2}}{3}$$ $$\frac{1}{3}(4 + \sqrt{2})$$

Therefore, the answer is $\fbox{(D)}$

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