Chapter 11.6, Problem 15WE

To determine

To calculate:

Value of $S_1,S_2,S_3,S_4,S_5$ and $S$ .

Answer to Problem 15WE

Value of $S$ is $\frac{1}{2}$ and $S_1=\frac{3}{4},S_2=0.37,S_3=0.56,S_4=0.46,S_5=0.51$

Explanation of Solution

Give information:

  $\frac{3}{4}-\frac{3}{8}+\frac{3}{16}-\frac{3}{32}+\mathrm{...}$

Calculation:

Given is $\frac{3}{4}-\frac{3}{8}+\frac{3}{16}-\frac{3}{32}+\mathrm{...}$ ,

  $\begin{array}{l}{S}_1=\frac{3}{4}\\ S_2=\frac{3}{4}-\frac{3}{8}\\ =0.375\\ S_3=\frac{3}{4}-\frac{3}{8}+\frac{3}{16}\\ =0.5625\\ S_4=\frac{3}{4}-\frac{3}{8}+\frac{3}{16}-\frac{3}{32}\\ S_4=0.46\\ S_5=\frac{3}{4}-\frac{3}{8}+\frac{3}{16}-\frac{3}{32}+\frac{3}{64}\\ S_5=0.51\end{array}$

The formula for sum of an infinite geometric series is,

  $S=\frac{s_1}{1-r}$

Where,

  $r=$ common ratio.

  $s_1=$ first term.

Since $-1<r<1$

Here common ratio is $=\frac{\left(\frac{-3}{8}\right)}{\left(\frac{3}{4}\right)}=-0.5$

Then put the value in equation of $S$ ,

  $\begin{array}{l}S=\frac{3/4}{1-\left(-0.5\right)}\\ S=\frac{3/4}{1+\left(0.5\right)}\\ S=\frac{3/4}{1.5}\\ S=\frac{1}{2}\end{array}$

Therefore, Value of $S$ is $\frac{1}{2}$ .