Step-Up Rationalization II

Algebra Level 3

4 ( 5 + 1 ) ( 5 4 + 1 ) ( 5 8 + 1 ) ( 5 16 + 1 ) \dfrac{4}{(\sqrt{5}+1)(\sqrt[4]{5}+1)(\sqrt[8]{5}+1)(\sqrt[16]{5}+1)}

Let the above expression be denoted as x x , find the value of ( x + 1 ) 48 (x+1)^{48} .

Image Credit: Wikimedia nTH root


The answer is 125.

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

2 solutions

Danish Ahmed
Mar 12, 2015

x = 4 ( 5 16 1 ) ( 5 + 1 ) ( 5 4 + 1 ) ( 5 8 + 1 ) ( 5 16 + 1 ) ( 5 16 1 ) = . . . = 5 16 1 x = \dfrac{4(\sqrt[16]{5}-1)}{(\sqrt{5}+1)(\sqrt[4]{5}+1)(\sqrt[8]{5}+1)(\sqrt[16]{5}+1)(\sqrt[16]{5}-1)}=...=\sqrt[16]{5}-1

( x + 1 ) 48 = 5 3 = 125 (x+1)^{48}=5^{3}=125

Putting in another way, let a = 5 16 . a 16 = 5 Multiplying numerator and denominator by (a-1) x = 4 ( a 1 ) ( a 8 + 1 ) ( a 4 + 1 ) ( a 2 + 1 ) ( a + 1 ) ( a 1 ) x = 4 ( a 1 ) a 16 1 = a 1. ( x + 1 ) 48 = ( a 1 + 1 ) 48 = ( 5 16 ) 48 = 125 \large a=\sqrt[16]5 .~~~~\implies a^{16}=5\\\text{ Multiplying numerator and denominator by (a-1)}\\\large x=\dfrac{4(a-1)}{(a^8+1)(a^4+1)(a^2+1)(a+1)(a-1)}\\\large x=\dfrac{4(a-1)}{a^{16}-1}=a-1.\\\large \therefore (x+1)^{48}=(a-1+1)^{48}=(\sqrt[16]5)^{48}=125
This is the SAME problem that had appeared some time back.

Actually that was my problem some time ago.

William Isoroku - 6 years ago

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...