Question

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Answer 1

(2^2001 + 2^1999) / (2^2000 + 2^1998)

…Taking 2^1998 common in the numerator…

=2^1998 (2^(2001–1998) + 2^(1999–1998)) / (2^2000 + 2^1998)

{as, a^m/ a^n = a^(m-n) }

=2^1998 (2^3 + 2^1) / (2^2000 + 2^1998)

=2^1998 (8 + 2) / (2^2000 + 2^1998)

=2^1998 (10) / (2^2000 + 2^1998)

…Taking 2^1998 common in the denominator…

=2^1998 (10) / 2^1998 (2^(2000–1998) + 2^(1998–1998))

=2^1998 (10) / 2^1998 (2^2 + 2^0)

=2^1998 (10) / 2^1998 (4+1)

{as a^0 = 1 }

=2^1998 (10) / 2^1998 (5)

…As 2^1998 is common both to the numerator and the denominator, it may be cancelled from both.

=10 / 5

= 2


$hope \: that \: this \: answer \: will \: help \: u$


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Answer 2

SO FROM SOLLUTION THE RIGHT OPTION WILL BE ( 1 )

some basics to solve this question :

1/a = a^-1
a^n + a^n = a^n ( 1 + 1 )

a^n / a^m = a^n-m

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