Question

$\frac{16\times 2^{n+1}-4\times 2^n}{16\times 2^{n+2}-2\times 2^{n+2}}$ equals

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

Answer 1

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

$=\frac{16\times 2^{n+1}-4\times 2^n}{16\times 2^{n+2}-2\times 2^{n+2}}$

We know, $a^{m+l}=a^m\times a^l$

Thus, $2^{n+1}=2^n\times 2^1$

$2^{n+2}=2^n\times 2^2$

$=\frac{16\times 2^n\times 2-4\times 2^n}{16\times 2^n\times 2^2-2\times 2^n\times 2^2}$

$=\frac{2^n(16\times 2-4)}{2^n(116\times 4-2\times 4)}=\frac{32-4}{64-8}=\frac{28}{56}=\frac{1}{2}$