Chapter 12, Problem 7CT

To determine

To find: The common ratio and the sum of the first $n$ terms.

Answer to Problem 7CT

The sum of $n$ terms of the sequence is $s_n=\frac{2}{3}\left(1-4^n\right)$ .

Explanation of Solution

Given information:

The given sequence is $\left\{-\frac{1}{2}\cdot 4^n\right\}$ .

Calculation:

Substitute the values of $1,2,3$ and $4$ for $n$ in the series $\left\{-\frac{1}{2}\cdot 4^n\right\}$ .

Thus, $a_1=-2$ , $a_2-8$ , $a_3=-32$ and $a_4=-128$ .

The sequence is $-2,-8,-32,-\mathrm{128......................}$

Calculate the difference between successive terms.

  $a_2-a_1=-8-\left(-2\right)=-6$

  $a_3-a_2=-32-\left(-8\right)=-24$

The difference between successive terms is not constant.

Substitute the value of $-2$ for $a_1$ and $4$ for $r$ to get the sum of $n$ terms.

  $\begin{array}{c}{s}_n=-2\cdot \frac{1-{\left(4\right)}^n}{1-\left(4\right)}\\ =\frac{2}{3}\left[1-{\left(4\right)}^n\right]\\ =\frac{2}{3}\left(1-4^4\right)\end{array}$

Therefore, the sum of $n$ terms of the sequence is $s_n=\frac{2}{3}\left(1-4^n\right)$ .