Question

Find the value of $\frac{16\times 2^{n+1}-4\times 2^n}{16\times 2^{n+2}-2\times 2^{n+2}}.$

• A. $\frac{1}{4}$
• B. $\frac{1}{2}$
• C. 1
• D. 2

Answer 1

The correct option is B $\frac{1}{2}$

Let u$=\frac{16\times 2^{n+1}-4\times 2^n}{16\times 2^{n+2}-2\times 2^{n+2}}$
We know that $a^{m+l}=a^m\times a^l$
Hence
u $=\frac{16\times 2^n\times 2-4\times 2^n}{16\times 2^n\times 2^2-2\times 2^n\times 2^2}$
u=$\frac{2^n(16\times 2-4)}{2^n(16\times 4-2\times 4)}$
$=\frac{32-4}{64-8}$
$=\frac{28}{56}$

u$=\frac{1}{2}$