Question

David is trying to solve the expression :
\[\frac{(4)^2 \times 2^{(n+1)} - 4 \times 2^n}{(4)^2 \times 2^{(n+2)} - 2 \times 2^{(n+2)}}\]
And you help him to do the same and finally arrive at the answer with correct to one decimal which would be - (Note:- DO NOT include spaces in your answer)

Answer 1

Correct Answer: 0.5

Let's simplify the given expression step-by-step.
Given expression:
\[\frac{(4)^2 \times 2^{(n+1)} - 4 \times 2^n}{(4)^2 \times 2^{(n+2)} - 2 \times 2^{(n+2)}}\]
Step 1: Simplify the Numerator
The numerator is:
\[(4)^2 \times 2^{(n+1)} - 4 \times 2^n\]
Since \((4)^2 = 16\), the numerator becomes:
\[16 \times 2^{(n+1)} - 4 \times 2^n\]
Step 2: Simplify the Denominator
The denominator is:
\[(4)^2 \times 2^{(n+2)} - 2 \times 2^{(n+2)}\]
Again, since \((4)^2 = 16\), the denominator becomes:
\[16 \times 2^{(n+2)} - 2 \times 2^{(n+2)}\]
\[= (16 - 2) \times 2^{(n+2)}\]
\[= 14 \times 2^{(n+2)}\]
Step 3: Combine and Simplify
Now let's plug the simplified numerator and denominator back into the expression:
\[\frac{16 \times 2^{(n+1)} - 4 \times 2^n}{14 \times 2^{(n+2)}}\]
Factor out \(2^n\) from the numerator:
\[= \frac{2^n (16 \times 2 - 4)}{14 \times 2^{(n+2)}}\]
\[= \frac{2^n (32 - 4)}{14 \times 2^{(n+2)}}\]
\[= \frac{2^n (28)}{14 \times 2^{(n+2)}}\]
Factor out \(2^{n+2}\) from the denominator:
\[= \frac{28 \times 2^n}{14 \times 2^{(n+2)}}\]
Since \(2^{(n+2)} = 4 \times 2^n\):
\[= \frac{28 \times 2^n}{14 \times 4 \times 2^n}\]
Cancel \(2^n\):
\[= \frac{28}{14 \times 4}\]
Simplify:
\[= \frac{28}{56}\]
Final Answer:
\[= \frac{1}{2}\]
Therefore, the value is 0.5.