Chapter 9.4, Problem 6E

Solution

To state: The sum $5-15+45-135+\cdots$ using the summation notation.

The resultant summation notation is ${\displaystyle \sum _{k=1}^{\infty }5{\left(-3\right)}^{k-1}}$ .

Given information:

The given sum is $5-15+45-135+\cdots$ .

Explanation:

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

Now consider the given sum,

  $5-15+45-135+\cdots$

The first term is $a_1=5$ .

The common ratio ( r ) is:

  $\begin{array}{l}r=\frac{-15}{5},r=-3\\ r=\frac{45}{-15},r=-3\\ r=\frac{-135}{45},r=-3\end{array}$

The nth term of a geometric sequence is:

  $a_n=a_1\cdot r^{n-1}$

Substitute the values of

  $a_1\text{ and }r$ to get the value of n ,

  $a_n=5\cdot {\left(-3\right)}^{n-1}$

The lower limit of the summation is 1 and there is no upper limit for the summation. Then the summation notation replacing n with k is: ${\displaystyle \sum _{k=1}^{\infty }5{\left(-3\right)}^{k-1}}$

The summation notation of the series is ${\displaystyle \sum _{k=1}^{\infty }5{\left(-3\right)}^{k-1}}$ .