Question

Simplify: $\frac{16\times 2^{n+1}-8\times 2^n}{16\times 2^{n+2}-4\times 2^{n+1}}$

Answer 1

We have:
$\frac{16\times 2^{n+1}-8\times 2^n}{16\times 2^{n+2}-4\times 2^{n+1}}$

⇒ $\frac{2^4\times 2^{n+1}-2^3\times 2^n}{2^4\times 2^{n+2}-2^2\times 2^{n+1}}$

⇒ $\frac{2^3\times \left(2^{n+2}-2^n\right)}{2^3\times \left(2^{n+3}-2^n\right)}$

⇒ $\frac{2^n\times 2^2-2^n}{2^n\times 2^3-2^n}$

⇒ $\frac{2^n\left(2^2-1\right)}{2^n\left(2^3-1\right)}=\frac{4-1}{8-1}=\frac{3}{7}$