Question

plz solve it
The one who will solve,I will follow him/her.........​

Answer 1

$\frac{16 \times 2^{n+1}-8 \times 2^n}{16 \times 2^{n+2}-4 \times 2^{n+1}} \\ \\ =\frac{2^4 \times 2^{n+1}-2^3 \times 2^n}{2^4 \times 2^{n+2}-2^2 \times 2^{n+1}} \\ \\ =\frac{2^{n+5}-2^{n+3}}{2^{n+6}-2^{n+3}} \\ \\ =\frac{2^{n+3}(2^2-1)}{2^{n+3}(2^3-1)} \\ \\ =\frac{2^2-1}{2^3-1} \\ \\ = \frac{4-1}{8-1}\\ \\ = \frac{3}{7}$

Answer 2

$\large{ \mathfrak{ \pmb{ \red{Question : }}}}$

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

$\large{\mathfrak{ \pmb{ \red{Step - by - step \: explanation : }}}}$

$\begin{gathered}\frac{16 \times 2^{n+1}-8 \times 2^n}{16 \times 2^{n+2}-4 \times 2^{n+1}} \\ \\ ⇢ \: \frac{2^4 \times 2^{n+1}-2^3 \times 2^n}{2^4 \times 2^{n+2}-2^2 \times 2^{n+1}} \\ \\ ⇢ \: \frac{2^{n+5}-2^{n+3}}{2^{n+6}-2^{n+3}} \\ \\⇢ \: \frac{2^{n+3}(2^2-1)}{2^{n+3}(2^3-1)} \\ \\ ⇢ \: \frac{2^2-1}{2^3-1} \\ \\ ⇢ \: \frac{4-1}{8-1}\\ \\ ⇢ \: { \purple{\frac{3}{7}}}\end{gathered}$