Question

Simplify the following:

(i)

(ii)

(iii)

(iv)

Answer 1

(i)
$$\begin{gathered} \frac{3^n\times 9^{n+1}}{3^{n-1}\times 9^{n-1}} \\ =\frac{3^n\times {\left(3^2\right)}^{\left(n+1\right)}}{3^{n-1}\times {\left(3^2\right)}^{n-1}} \\ =\frac{3^n\times 3^{2n+2}}{3^{n-1}\times 3^{2n-2}} \end{gathered}$$
$$\begin{gathered} =\frac{3^{n+2n+2}}{3^{n-1+2n-2}} \\ =\frac{3^{3n+2}}{3^{3n-3}} \\ =3^{3n+2-3n+3} \\ =3^5 \\ =243 \end{gathered}$$

(ii)
$$\begin{gathered} \frac{5\times {25}^{n+1}-25\times 5^{2n}}{5\times 5^{2n+3}-{25}^{n+1}} \\ =\frac{5\times {\left(5^2\right)}^{n+1}-\left(5^2\right)\times 5^{2n}}{5\times 5^{2n+3}-{\left(5^2\right)}^{n+1}} \\ =\frac{5\times \left(5^{2n+2}\right)-\left(5^2\right)\times 5^{2n}}{5\times 5^{2n+3}-\left(5^{2n+2}\right)} \\ =\frac{5^{1+2n+2}-5^{2+2n}}{5^{1+2n+3}-5^{2n+2}} \end{gathered}$$
$$\begin{gathered} =\frac{5^{2n+3}-5^{2+2n}}{5^{2n+4}-5^{2n+2}} \\ =\frac{5^{2+2n}\left(5-1\right)}{5^{2n+2}\left(5^2-1\right)} \\ =\frac{4}{24} \\ =\frac{1}{6} \end{gathered}$$

(iii)
$$\begin{gathered} \frac{5^{n+3}-6\times 5^{n+1}}{9\times 5^n-2^2\times 5^n} \\ =\frac{5^{n+1}\left(5^2-6\right)}{5^n\left(9-2^2\right)} \\ =\frac{5^n\times 5\times \left(25-6\right)}{5^n\left(9-4\right)} \\ =\frac{5\times 19}{5} \\ =19 \end{gathered}$$

(iv)
$$\begin{gathered} \frac{6{\left(8\right)}^{n+1}+16{\left(2\right)}^{3n-2}}{10{\left(2\right)}^{3n+1}-7{\left(8\right)}^n} \\ =\frac{6{\left(2^3\right)}^{n+1}+16{\left(2\right)}^{3n-2}}{10{\left(2\right)}^{3n+1}-7{\left(2^3\right)}^n} \\ =\frac{6\left(2^{3n+3}\right)+16{\left(2\right)}^{3n-2}}{10{\left(2\right)}^{3n+1}-7\left(2^{3n}\right)} \end{gathered}$$
$$\begin{gathered} =\frac{6\times 2^{3n}\left(2^3\right)+16{\left(2\right)}^{3n}{2}^{-2}}{10{\left(2\right)}^{3n}\left(2^1\right)-7\left(2^{3n}\right)} \\ =\frac{2^{3n}\left(48+4\right)}{2^{3n}\left(20-7\right)} \\ =\frac{52}{13} \\ =4 \end{gathered}$$