Chapter 8.5, Problem 5E

To determine

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

Answer to Problem 5E

  $1,-4,-9,-14,-19,-24$ .

Explanation of Solution

Given:

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

  $f\left(0\right)=4$

Calculation:

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

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

  $\begin{array}{l}f\left(n\right)=2f\left(n-1\right)\{\text{given sequence}}\\ f\left(1\right)=2f\left(1-1\right)\{\text{put }n=1\text{}}\\ f\left(1\right)=2f\left(0\right)\\ f\left(1\right)=2\left(4\right)=8\{\text{substitutes }f\left(0\right)=4\text{}}\\ f\left(2\right)=2f\left(2-1\right)\{\text{put }n=2\text{}}\\ f\left(2\right)=2f\left(1\right)\\ f\left(2\right)=2\left(8\right)=16\{\text{substitutes }f\left(1\right)=8\text{}}\end{array}$

  $\begin{array}{l}f\left(3\right)=2f\left(3-1\right)\{\text{put }n=3\text{}}\\ f\left(3\right)=2f\left(2\right)\\ f\left(3\right)=2\left(16\right)=32\{\text{substitutes }f\left(2\right)=16\text{}}\\ f\left(4\right)=2f\left(4-1\right)\{\text{put }n=4\text{}}\\ f\left(4\right)=2f\left(3\right)\\ f\left(4\right)=2\left(32\right)=64\{\text{substitutes }f\left(3\right)=32\text{}}\\ f\left(5\right)=2f\left(5-1\right)\{\text{put }n=4\text{}}\\ f\left(5\right)=2f\left(4\right)\\ f\left(5\right)=2\left(64\right)=128\{\text{substitutes }f\left(4\right)=64\text{}}\end{array}$

Hence, the first six terms of the given sequence are $4,8,16,32,64,128$ .