Question

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

Answer 1

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

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

$=\frac{2^3\times 2^1\times 2^{4+1}-2^{4+2}}{2^{4+2}\times 14}$ $[\because a^m\times a^4=a^{m+4}]$

$=\frac{2^3\times 2^{n+2}-2^{4+2}}{2^{4+2}\times 14}$

$=\frac{2^{4+2}(8-1)}{2^{4+2}\times 14}$

$=\frac{7}{14}$

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

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