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Let x =                                                            What is \((x+1)^{48}\)

                               4

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

 Dec 31, 2020
 #1
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This question was posted before and the answer should have been 125

 Dec 31, 2020
 #2
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I just entered it all in a calculator....    answer is 125

   Really no solving involved.....just enter it into a calculator.....to find 'x' then add 1   And 48th power it.

 Dec 31, 2020
edited by Guest  Dec 31, 2020
 #3
avatar+112823 
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I think the point of this question is that it is meant to be done without a calculator.

 

\(Let\;\;x=\frac{4}{(\sqrt{5}+1)(\sqrt[4]{5}+1)(\sqrt[8]{5}+1) (\sqrt[16]{5}+1)}\qquad find\;\;(x+1)^{48}\)

 

 

Let    \(y=5^{1/16}\)

 

\((\sqrt{5}+1)(\sqrt[4]{5}+1)(\sqrt[8]{5}+1) (\sqrt[16]{5}+1)\\ =(y^8+1)(y^4+1)(y^2+1)(y+1)\\ =y^{12}+y^{11}+y^{10}+y^9 ..... y^1+1\\ \text{This is the sum of a GP with a= 1, r=y, and n=16}\\ =\frac{a(r^n-1)}{r-1}=\frac{y^{16}-1}{y-1}=\frac{5-1}{y-1}=\frac{4}{y-1}\)

 

so

\(x=\frac{4}{\frac{4}{y-1}}=y-1\\~\\ (x+1)^{48}\\=(y-1+1)^{48}\\=y^{48}\\=(5^{1/16})^{48}\\=5^3\\=125\)

 

 

 

LaTex:

(\sqrt{5}+1)(\sqrt[4]{5}+1)(\sqrt[8]{5}+1) (\sqrt[16]{5}+1)\\
=(y^8+1)(y^4+1)(y^2+1)(y+1)\\
=y^{12}+y^{11}+y^{10}+y^9 ..... y^1+1\\
\text{This is the sum of a GP with a= 1,  r=y, and n=16}\\
=\frac{a(r^n-1)}{r-1}=\frac{y^{16}-1}{y-1}=\frac{5-1}{y-1}=\frac{4}{y-1}

 

x=\frac{4}{\frac{4}{y-1}}=y-1\\~\\
(x+1)^{48}\\=(y-1+1)^{48}\\=y^{48}\\=(5^{1/16})^{48}\\=5^3\\=125

 Dec 31, 2020

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