If $S_n = 6 - 2^{n+1}/3^{n-1}$, can we show that we have a GP?

If $S_n = 6 - 2^{n+1}/3^{n-1}$, can we show that we have a GP?

Question

The sum of the first

$n$

terms of a series is given by the expression

$$\begin{equation*} 6 - \frac{2^{n+1}}{3^{n-1}}. \end{equation*}$$

By finding an expression for the $n$th term of the series, or otherwise, show that this is a geometric series, and state the values of the first term and the common ratio.

Solution

Let the sequence that we are summing be

$(a_n)$

for

$n \ge 1$

, so that

$$\begin{equation*} a_1 + \dots b + a_n = 6 - \frac{2^{n+1}}{3^{n-1}} \end{equation*}$$

for every

$n \ge 1$

. In particular,

$$\begin{equation*} a_1 = 6 - \frac{2^2}{3^0} = 6 - 4 = 2, \end{equation*}$$

and, for every

$n \ge 2$

,

$$\begin{align*} a_{n} = \left( a_1 + \dots b + a_{n} \right) - \left( a_1 + \dots b + a_{n-1} \right) &= \left( 6 - \frac{2^{n+1}}{3^{n-1}} \right) - \left( 6 - \frac{2^{n}}{3^{n-2}} \right) \\ &= \frac{2^{n}}{3^{n-2}} - \frac{2^{n+1}}{3^{n-1}} \\ &= \frac{2^{n}}{3^{n-2}} \left( 1 - \frac{2}{3} \right) \\ &= \frac{2^{n}}{3^{n-2}} \frac{1}{3} \\ &= \frac{2^{n}}{3^{n-1}} \\ &= 2 \left( \frac{2}{3} \right)^{n-1}. \end{align*}$$

Thus, $(a_n)$ is a geometric progression with first term $a_1 = 2$ and common ratio $\dfrac{2}{3}$.