Chapter 9.4, Problem 42E

To determine

To solve:Find a formula for the sum of the first n terms of the sequence $3,-\frac{9}{2},\frac{27}{4},-\frac{81}{8},\mathrm{.....}$ and verify it.

Explanation of Solution

Given:

  $3,-\frac{9}{2},\frac{27}{4},-\frac{81}{8},\mathrm{.....}$

Calculation:

In the given sequence $3,-\frac{9}{2},\frac{27}{4},-\frac{81}{8},\mathrm{.....}$ we can see that the sequence is a geometric progression with common ratio $-\frac{3}{2}$ .

The sum of first n terms of a G.P. is given by $S=a\left(\frac{1-r^n}{1-r}\right)whenr<1$ where a is the first term of the series and r is the common ratio.

Here, $a=3,r=-\frac{3}{2}$

We have,

  $\begin{array}{l}S=3-\frac{9}{2}+\frac{27}{4}-\frac{81}{8}+\mathrm{.....}uptonterms\\ =>S=3\left(\frac{1-{\left(-\frac{3}{2}\right)}^n}{1-\left(-\frac{3}{2}\right)}\right)\\ =>S=3\left(\frac{1-{\left(-\frac{3}{2}\right)}^n}{1+\frac{3}{2}}\right)\\ =>S=\frac{6}{5}\left(1-{\left(-\frac{3}{2}\right)}^n\right)-------(1)\end{array}$

To verify our result, let us find the sum of the first two terms of the sequence manually and also by using the formula and examine if both are same or not.

Sum of first two terms $=\left(3-\frac{9}{2}\right)=\frac{6-9}{2}=-\frac{3}{2}$

Now, in (1) putting $n=2$ we get:

  $\begin{array}{l}=\frac{6}{5}\left(1-{\left(-\frac{3}{2}\right)}^2\right)\\ =\frac{6}{5}\left(1-\frac{9}{4}\right)\\ =\frac{6}{5}\left(\frac{4-9}{4}\right)\\ =\frac{6}{5}\left(\frac{-5}{4}\right)\\ =-\frac{3}{2}\end{array}$

Hence, our formula is verified.