Chapter 9, Problem 28MCQ

Solution

To find the amount that will be received in tenth month.

The amount in the tenth month will be $104.

Given information:

Over the first four months someone receives $50, $52, $55 and $59.

Calculation:

Observe that each month the amount is increased by 1 than the amount increased in the previous month. Continuing in this way, the sequence can be written as $50, $52, $55, $59, $64, $70, $77, $85, $94, $104.

Therefore, the amount received in the tenth month will be $104.

Conclusion:

The amount in the tenth month will be $104.

a.

To write a formula to describe the sequence.

The formula for the sequence is $a_n=a_{n-1}+n$ , where $a_1=50$ .

Given information:

Over the first four months someone receives $50, $52, $55 and $59.

Calculation:

The terms of the sequence can be written as follows:

  $\begin{array}{l}{a}_1=50\\ a_2=52=50+2=a_1+2\\ a_3=55=52+3=a_2+3\\ a_4=59=55+4=a_3+4\end{array}$

Therefore, the nth term of the sequence can be written as $a_n=a_{n-1}+n$ , where $a_1=50$ .

Conclusion:

The formula for the sequence is $a_n=a_{n-1}+n$ , where $a_1=50$ .

b.

To identify the formula as recursive or explicit.

The formula $a_n=a_{n-1}+n$ , where $a_1=50$ is recursive.

Given information:

Over the first four months someone receives $50, $52, $55 and $59.

Calculation:

From part (a), the formula for the sequence is $a_n=a_{n-1}+n$ , where $a_1=50$ .

Observe that starting from second term, each term is dependent on its previous term. So, the formula is recursive.

Conclusion:

The formula $a_n=a_{n-1}+n$ , where $a_1=50$ is recursive.

c.

To explain the difference between a recursive and explicit formula.

Given information:

Over the first four months someone receives $50, $52, $55 and $59.

Calculation:

From part (a), the formula for the sequence is $a_n=a_{n-1}+n$ , where $a_1=50$ .

Observe that starting from second term, each term is dependent on its previous term. So, the formula is recursive. On the other hand, in an explicit formula the terms are not dependent on the previous terms, and each term can be found by using the value of $n$ .