Chapter 12.1, Problem 58E

To determine

To calculate: The sum of first 72 terms of the sequence $6,8,2,\dots$ where $a_n=a_{n-1}-a_{n-2}$ .

Answer to Problem 58E

The sum of first 72 terms of the sequence is0.

Explanation of Solution

Given information:

The sequence $6,8,2,\dots$ where $a_n=a_{n-1}-a_{n-2}$ .

Calculation:

Consider the sequence $6,8,2,\dots$ where $a_n=a_{n-1}-a_{n-2}$ .

Compute the next few terms of the sequence with help of general formula.

Substitute n as 4 in $a_n=a_{n-1}-a_{n-2}$ ,

  $\begin{array}{c}{a}_4=a_3-a_2\\ =2-8\\ =-6\end{array}$

Substitute n as 5in $a_n=a_{n-1}-a_{n-2}$ ,

  $\begin{array}{c}{a}_5=a_4-a_3\\ =-6-2\\ =-8\end{array}$

Substitute n as 6in $a_n=a_{n-1}-a_{n-2}$ ,

  $\begin{array}{c}{a}_6=a_5-a_4\\ =-8+6\\ =-2\end{array}$

Substitute n as 7in $a_n=a_{n-1}-a_{n-2}$ ,

  $\begin{array}{c}{a}_7=a_6-a_5\\ =-2+8\\ =6\end{array}$

Therefore, the sequence is $6,8,2,-6,-8,-2,6,\dots$ .

Now, observe the sequence first three terms are positive and next three terms are negative of it.

The series is repeated after 6 terms.

When first six terms are summed the result is zero as each term cancel its negative.

For the sum of first 72 terms, so there will be 12 groups of 6 terms in which first three terms are positive and next three terms are negative of it.

So each group sum will be 0. Therefore, sum of first 72 terms will be 0.

Thus, the sum of first 72 terms of the sequence $6,8,2,\dots$ is 0.