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

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 is0.5.