Solve {l}{a_{n}=3^n+4/3^n}{n=1}{a=2}{text{Solvefor}btext{where}}{b=a}

$$\left. \begin{array} { l } { a_{n} = \frac{3 ^ {n + 4}}{3 ^ {n}} }\\ { n = 1 }\\ { a = 2 }\\ { \text{Solve for } b \text{ where} } \\ { b = a } \end{array} \right.$$

Solve for a_n, n, a, b

$$b=2$$

Solution Steps

Consider the first equation. Insert the known values of variables into the equation.

$$a_{n}=\frac{3^{1+4}}{3^{1}}$$

Add 1 and 4 to get 5.

$$a_{n}=\frac{3^{5}}{3^{1}}$$

Calculate 3 to the power of 5 and get 243.

$$a_{n}=\frac{243}{3^{1}}$$

Calculate 3 to the power of 1 and get 3.

$$a_{n}=\frac{243}{3}$$

Divide 243 by 3 to get 81.

$$a_{n}=81$$

The system is now solved.

$$a_{n}=81 n=1 a=2 b=2$$