Theory of Intelligence
11TH•Mathematics•Sequences And Series•Question 7

Overview

Show that $\sqrt{4} \cdot \sqrt[4]{16} \cdot \sqrt[8]{64} \cdot \sqrt[16]{256} \cdots \infty = 16$.

[ P=\sqrt{4}\cdot\sqrt[4]{16}\cdot\sqrt[8]{64}\cdot\sqrt[16]{256}\cdots =4^{1/2}\cdot 16^{1/4}\cdot 64^{1/8}\cdot 256^{1/16}\cdots ] [ = (2^2)^{1/2}\cdot(2^4)^{1/4}\cdot(2^6)^{1/8}\cdot(2^8)^{1/16}\cdots =2^{\sum 2n/2^n}=2^{2\sum n/2^n}=2^{2\cdot 2}=2^4=16 ]

Proved.

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