Chapter 9.4, Problem 1E

Solution

To state: The sum $-7-1+5+11+\cdots +53$ using the summation notation.

The resultant summation notation is ${\displaystyle \sum _{k=1}^{11}6k-13}$ .

Given information:

The given sum is $-7-1+5+11+\cdots +53$ .

Explanation:

When the terms of a sequence are added together, the resulting expression is a series.

Now consider the given sum,

  $-7-1+5+11+\cdots +53$

The first term is $a_1=-7$ .

The common difference ( d ) is:

  $\begin{array}{c}-1-\left(-7\right)=6\\ 5-\left(-1\right)=6\\ 11-5=6\end{array}$

The nth term of an arithmetic sequence is:

  $a_n=a_1+\left(n-1\right)d$

Substitute the values of $a_1\text{ and }d$ to get the value of n ,

  $\begin{array}{c}{a}_n=-7+\left(n-1\right)6\\ a_n=-7+6n-6\\ a_n=6n-13\end{array}$

The upper limit depends on the last term so put $a_n=53$ and solve for n ,

  $\begin{array}{l}53=6n-13\\ 66=6n\\ 11=n\end{array}$

The lower limit of the summation is 1 and the upper limit of the summation is 11.

The summation notation of the series is ${\displaystyle \sum _{k=1}^{11}6k-13}$ .