Question

CAN SOMEONE PLEASE HELP

Answer 1

Answer:

$\sf \dfrac{1}{2^n+1}$

Step-by-step explanation:

Given expression:

$\sf \dfrac{2^n-1}{4^n-1}$

Rewrite $\sf 4^n$:   $\sf 4^n=(2^2)^n=2^{2n}=(2^n)^2$

Rewrite 1:  1 = 1²

$\sf \implies \dfrac{2^n-1}{(2^n)^2-1^2}$

Apply difference of two square formula to the denominator

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

$\sf \implies \dfrac{2^n-1}{(2^n+1)(2^n-1)}$

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

$\sf \implies \dfrac{1}{2^n+1}$

Answer 2

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

Here's the solution ~

$\qquad \sf \dashrightarrow \: \dfrac{2 {}^{n} - 1}{ {4}^{n} - 1}$

$\qquad \sf \dashrightarrow \: \dfrac{2 {}^{n} - 1}{ {( {2}^{2}) }^{n} - 1}$

$\qquad \sf \dashrightarrow \: \dfrac{2 {}^{n} - 1}{ {( {2}^{n}) }^{2} - 1}$

Now we will use this identity in denominator

[ a² - b² = (a + b)(a - b) ]

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

$\qquad \sf \dashrightarrow \: \dfrac{ 1}{ {( {2}^{n} }^{} + 1) }$