Question

CAN SOMEONE PLEASE HELP

Answer 1

`\qquad\qquad\huge\underline{{\sf Answer}}♨`

Here's the solution ~

`\qquad \sf  \dashrightarrow \: \frac{2 {}^(n) - 1}{ {4}^(n) - 1}`

`\qquad \sf  \dashrightarrow \: \frac{2 {}^(n) - 1}{ {( {2}^(2)) }^(n) - 1}`

`\qquad \sf  \dashrightarrow \: \frac{2 {}^(n) - 1}{ {( {2}^(n)) }^(2) - 1}`

Now we will use this identity in denominator

[ a² - b² = (a + b)(a - b) ]

`\qquad \sf  \dashrightarrow \: \frac{ \cancel{(2 {}^(n) - 1)}}{ {( {2}^(n) }^{} + 1) \cancel{( {2}^(n) - 1)}}`

`\qquad \sf  \dashrightarrow \: \frac{ 1}{ {( {2}^(n) }^{} + 1) }`

Answer 2

`\sf (1)/(2^n+1)`

Explanation:

Given expression:

`\sf (2^n-1)/(4^n-1)`

Rewrite
`\sf 4^n`:
`\sf 4^n=(2^2)^n=2^(2n)=(2^n)^2`

Rewrite 1: 1 = 1²

`\sf \implies (2^n-1)/((2^n)^2-1^2)`

Apply difference of two square formula to the denominator

`\sf a^2-b^2=(a+b)(a-b)`

`\sf \implies (2^n-1)/((2^n+1)(2^n-1))`

Cancel out the common factor
`\sf 2^n-1` :

`\sf \implies (1)/(2^n+1)`