Chapter 12.3, Problem 47E

To determine

Calculate the sum of the given geometric sequence $1+3+9+\mathrm{.......}+2187$ .

Answer to Problem 47E

Sum of the given geometric sequence $1+3+9+\mathrm{.......}+2187$ is $S_8=3280$

Explanation of Solution

Given information:

It is given in the question that, geometric sequence $1+3+9+\mathrm{.......}+2187$ .

Formula used:

  $S_n=a\frac{1-r^n}{1-r}\&a_n=a{r}^{n-1}$

Where,

  $\begin{array}{l}aisthe{1}^{st}term\\ nisthenumberofterm\\ risthecommonratio\\ S_{n}isthepartialsumofnthterm\end{array}$

Calculation:

The given geometric sequence $1+3+9+\mathrm{.......}+2187$ means,

  $\begin{array}{l}{a}_1=1\Rightarrow a=1\\ a_2=3\Rightarrow ar=3\Rightarrow r=\frac{3}{a}\Rightarrow r=\frac{3}{1}\Rightarrow r=3\end{array}$

So, $a=1\&r=3$

Let $2187$ is the $nth$ term $\Rightarrow a_n=2187$

By using the formula $a_n=a{r}^{n-1}$

  $\begin{array}{l}{a}_n=2187\\ 1*{\left(3\right)}^{n-1}=2187\\ {\left(3\right)}^{n-1}={\left(3\right)}^7\end{array}$

By equating

  $n-1=7\Rightarrow n=8$

By using the formula $S_n=a\frac{1-r^n}{1-r}$

  $\begin{array}{l}{S}_8=1\frac{1-{\left(3\right)}^8}{1-3}\\ =1\left(\frac{1-6561}{1-3}\right)\\ =1\left(\frac{-6560}{-2}\right)\\ =3280\end{array}$