Chapter 9, Problem 30RE

Solution

To state: The first six terms and the 12^th term of the sequence:

  $b_1=5\text{and}{b}_k=2b_{k-1}\text{for}k\ge 2$ .

The resultants first six terms are $5,10,20,40,80,160$ and the 12^th term is 10,240.

Given information:

The given sequence is: $b_1=5\text{and}{b}_k=2b_{k-1}\text{for}k\ge 2$ .

Explanation:

Consider the given sequence,

  $b_k=2b_{k-1}\text{for}k\ge 2$

The first term of the sequence is: $b_1=5$

Now substitute 2 for k for the second term and simplify,

  $\begin{array}{c}{b}_2=2b_{2-1}\\ =2b_1\\ =2\left(5\right)\\ =10\end{array}$

Now substitute 3 for k for the third term and simplify,

  $\begin{array}{c}{b}_3=2b_{3-1}\\ =2b_2\\ =2\left(10\right)\\ =20\end{array}$

Similarly, find the remaining 3 terms,

  $\begin{array}{c}{b}_3=2b_{3-1}\\ =2b_2\\ =2\left(20\right)\\ =40\\ b_5=2b_{5-1}\\ =2b_4\\ =2\left(40\right)\\ =80\\ b_6=2b_{6-1}\\ =2b_5\\ =2\left(80\right)\\ =160\end{array}$

The first six terms are $5,10,20,40,80,160$ .

The sequence is a geometric sequence and each term is the double of the previous term, it means the common ratio is equal to 2.

Then, the 12^th term is:

  $\begin{array}{c}{b}_n=b_1{\left(r\right)}^{n-1}\\ b_{12}=5{\left(2\right)}^{12-1}\\ =5{\left(2\right)}^{11}\\ =5\left(2048\right)\\ =10,240\end{array}$

Therefore, the first six terms are $5,10,20,40,80,160$ and the 12^th term is 10,240.