Daizy14idn
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Sagot :

DᴀʀᴋPᴀʀᴀᴅᴏxidn

[tex]\qquad\qquad\huge\underline{{\sf Answer}}♨[/tex]

Here's the solution ~

[tex]\qquad \sf  \dashrightarrow \: \dfrac{2 {}^{n} - 1}{ {4}^{n} - 1} [/tex]

[tex]\qquad \sf  \dashrightarrow \: \dfrac{2 {}^{n} - 1}{ {( {2}^{2}) }^{n} - 1} [/tex]

[tex]\qquad \sf  \dashrightarrow \: \dfrac{2 {}^{n} - 1}{ {( {2}^{n}) }^{2} - 1} [/tex]

Now we will use this identity in denominator

[ a² - b² = (a + b)(a - b) ]

[tex]\qquad \sf  \dashrightarrow \: \dfrac{ \cancel{(2 {}^{n} - 1)}}{ {( {2}^{n} }^{} + 1) \cancel{( {2}^{n} - 1)}} [/tex]

[tex]\qquad \sf  \dashrightarrow \: \dfrac{ 1}{ {( {2}^{n} }^{} + 1) } [/tex]

Semsee45idn

Answer:

[tex]\sf \dfrac{1}{2^n+1}[/tex]

Step-by-step explanation:

Given expression:

[tex]\sf \dfrac{2^n-1}{4^n-1}[/tex]

Rewrite [tex]\sf 4^n[/tex]:   [tex]\sf 4^n=(2^2)^n=2^{2n}=(2^n)^2[/tex]

Rewrite 1:  1 = 1²

[tex]\sf \implies \dfrac{2^n-1}{(2^n)^2-1^2}[/tex]

Apply difference of two square formula to the denominator

[tex]\sf a^2-b^2=(a+b)(a-b)[/tex]

[tex]\sf \implies \dfrac{2^n-1}{(2^n+1)(2^n-1)}[/tex]

Cancel out the common factor [tex]\sf 2^n-1[/tex] :

[tex]\sf \implies \dfrac{1}{2^n+1}[/tex]

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