Chapter 8.5, Problem 6E

To determine

To write: the first six terms of the given sequence.

Answer to Problem 6E

  $10,5,\frac{5}{2},\frac{5}{4},\frac{5}{8},\frac{5}{16}$ .

Explanation of Solution

Given:

The given sequence is $f\left(n\right)=\frac{1}{2}f\left(n-1\right)$ .

  $f\left(0\right)=10$

Calculation:

According to the given sequence, $f\left(n\right)=\frac{1}{2}f\left(n-1\right)$ .

Therefore, the first six terms of the given sequence are $f\left(1\right),f\left(2\right),f\left(3\right),f\left(4\right),f\left(5\right)$ .

Now, to find the first six terms of the given sequence is given by

  $\begin{array}{l}f\left(n\right)=\frac{1}{2}f\left(n-1\right)\{\text{given sequence}}\\ f\left(1\right)=\frac{1}{2}f\left(1-1\right)\{\text{put }n=1\text{}}\\ f\left(1\right)=\frac{1}{2}f\left(0\right)\\ f\left(1\right)=\frac{1}{2}\left(10\right)=5\{\text{substitutes }f\left(0\right)=10\text{}}\\ f\left(2\right)=\frac{1}{2}f\left(2-1\right)\{\text{put }n=2\text{}}\\ f\left(2\right)=\frac{1}{2}f\left(1\right)\\ f\left(2\right)=\frac{1}{2}\left(5\right)=\frac{5}{2}\{\text{substitutes }f\left(1\right)=5\text{}}\end{array}$

  $\begin{array}{l}f\left(3\right)=\frac{1}{2}f\left(3-1\right)\{\text{put }n=3\text{}}\\ f\left(3\right)=\frac{1}{2}f\left(2\right)\\ f\left(3\right)=\frac{1}{2}\left(\frac{5}{2}\right)=\frac{5}{4}\{\text{substitutes }f\left(2\right)=\frac{5}{2}\text{}}\\ f\left(4\right)=\frac{1}{2}f\left(4-1\right)\{\text{put }n=4\text{}}\\ f\left(4\right)=\frac{1}{2}f\left(3\right)\\ f\left(4\right)=\frac{1}{2}\left(\frac{5}{4}\right)=\frac{5}{8}\{\text{substitutes }f\left(3\right)=\frac{5}{4}\text{}}\\ f\left(5\right)=\frac{1}{2}f\left(5-1\right)\{\text{put }n=4\text{}}\\ f\left(5\right)=\frac{1}{2}f\left(4\right)\\ f\left(5\right)=\frac{1}{2}\left(\frac{5}{8}\right)=\frac{5}{16}\{\text{substitutes }f\left(4\right)=\frac{5}{8}\text{}}\end{array}$

Hence, the first six terms of the given sequence are $10,5,\frac{5}{2},\frac{5}{4},\frac{5}{8},\frac{5}{16}$ .