Start by expanding \(\frac{3(2^{n+1}) - 4(2^{n-1})}{2^{n+1} - 2^n}\):
\(\frac{3 \times 2^n \times 2^1 - 2^2 \times 2^n \times 2^{-1}}{2^n \times 2 - 2^n}\)
NUMERATOR : 2\(^n\) ( 3\(^1\) X 2\(^1\) - 2\(^2\) X 2\(^{-1}\) )
--> 2\(^n\) ( 3 X 2 — 4 X \(\frac{1}{2}\) )
--> 2\(^n\) ( 6 - 2 )
--> 2\(^n\) (4)
DENOMINATOR : 2\(^n\) ( 2\(^1\) - 1 )
--> 2\(^n\) ( 2 - 1)
--> 2\(^n\)
⇒ \(\frac{2^n ( 4)}{2^n}\)
= 4
There is an explanation video available below.
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