Question

EVALUATE.............

Answer 1

Answer:

FIRSTLY TAKE 2^n common from numerator and as well as denominator,then

we get

= 2n+2n−12n+1−2n

take out 2^n common,we have

= (2n∗(1+1/2))(2n∗(2–1))

now upon cancellation of 2n on both numerator and denominator,we have

= (1+12)(2–1)

=321

=32

METHOD 2:-

PUT ANY VALUE OF n=0,1,2,3,………….

U SIMPLY GET

=32 .

please mark as brainliest

Answer 2

$\huge\bf\red{AnswEr:-}$

FIRSTLY TAKE 2^n common from numerator and as well as denominator,then

we get

= 2n+2n−12n+1−2n

take out 2^n common,we have

= (2n∗(1+1/2))(2n∗(2–1))

now upon cancellation of 2n on both numerator and denominator,we have

= (1+12)(2–1)

=321

=32