Algebra

Find the integer closest to $\frac { 1 }{ \sqrt [ 4 ]{ 5^{ 4 }+1 } -\sqrt [ 4 ]{ 5^{ 4 }-1 } }$

I wanted to post this as a question but found multiple answers Can anyone help me with this question? I think the answer might be $1$ OR $0$ OR $250$ which one is correct?

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Comments

Consider $\LARGE \frac {1}{ \sqrt[4]{A} - \sqrt[4]{B} }$

Multiply numerator and denominator by $\large \sqrt[4]{A} + \sqrt[4]{B}$ , apply difference of squares: $x^2 - y^2 = (x-y)(x+y)$ , we have

$\LARGE \frac { \sqrt[4]{A} + \sqrt[4]{B} }{ \sqrt[2]{A} - \sqrt[2]{B} }$

Now this time, multiply numerator and denominator by $\large \sqrt[2]{A} + \sqrt[2]{B}$

$\LARGE \frac { \left ( \sqrt[4]{A} + \sqrt[4]{B} \right ) \left ( \sqrt[2]{A} + \sqrt[2]{B} \right ) }{ A-B }$

Now set $A = 5^4 + 1, B = 5^4 - 1$ , the denominator equals to $2$

And because $A \approx 5^4, B \approx 5^4$ , then $\large \sqrt[4]{A} \approx 5, \sqrt[4]{B} \approx 5, \sqrt[2]{A} \approx 25, \sqrt[2]{B} \approx 25$

So the expression is appproximately close to $\frac {(5+5)(25+25)}{2} = \boxed{250}$

Alternatively, you can use binomial expansion, for $|x| <1$ , we have $(1 + x)^n \approx 1 + nx$

Because $\large \sqrt[4]{5^4 + 1} = \sqrt[4]{5^4 \left ( 1 + \frac {1}{5^4} \right ) } = 5 \cdot \sqrt[4]{ 1 + \frac {1}{5^4} }$ , likewise $\large \sqrt[4]{5^4 - 1} = 5 \cdot \sqrt[4]{ 1 - \frac {1}{5^4} }$

So the expression equals to $\large \left ( 5(1 + x)^n - 5(1-x)^n \right )^{-1}$

Substitution of $\large x = \frac {1}{5^4}, n = \frac {1}{4}$ gives $\large \frac {1}{10nx} = \boxed{250}$

Pi Han Goh - 6 years, 9 months ago

Beautiful answer...

Ankit Vijay - 6 years, 9 months ago

are you nuts??

abhinav jangir - 6 years, 1 month ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$