1024th root

Algebra Level 4

$\large \prod_{k=0}^{10} (2^{2^k}+1) =x^{1024}-1 \qquad, \qquad x = \ ?$

The answer is 4.

1 solution

Pieter-Jan Meuris
Apr 8, 2014

$\prod_{k=1}^{1024} (2^k+1)=\sum_{n=0}^m 2^n$

Now we search m:

$m= \sum_{p=0}^{10}2^p=2^{11}-1$

$(\sum_{n=0}^{2^{11}-1} 2^n)+1=...=2^{2^{11}}$

$\Rightarrow x=\sqrt[1024]{2^{2^{11}}}= \boxed{4}$

The problem is incorrect. The product $\prod_{k = 1}^{1024} (2^k + 1)$ is not equal to $4^{1024} - 1$ .

The intended product is probably $(2 + 1)(2^2 + 1)(2^4 + 1)(2^8 + 1) \dotsm (2^{2^{10}} + 1),$ because $\begin{aligned} &(2 + 1)(2^2 + 1)(2^4 + 1)(2^8 + 1) \dotsm (2^{2^{10}} + 1) \\ &= (2 - 1)(2 + 1)(2^2 + 1)(2^4 + 1)(2^8 + 1) \dotsm (2^{2^{10}} + 1) \\ &= (2^2 - 1)(2^2 + 1)(2^4 + 1)(2^8 + 1) \dotsm (2^{2^{10}} + 1) \\ &= (2^4 - 1)(2^4 + 1)(2^8 + 1) \dotsm (2^{2^{10}} + 1) \\ &= \dotsb \\ &= (2^{2^{10}} - 1)(2^{2^{10}} + 1) \\ &= 2^{2^{11}} - 1 \\ &= 4^{2^{10}} - 1. \end{aligned}$

Jon Haussmann - 7 years, 1 month ago

You're right.

Pieter-Jan Meuris - 7 years, 1 month ago

It should begin at k=0.

Hwang seung hwan - 7 years ago

1024th root

The answer is 4.

1 solution

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