Question

In Problems 71-82, find the sum of each sequence.

$\sum _{k=8}^{40}(-3k)$

Short answer

The sum of this sequence is -2376.

Step-by-step solution

Step 1. Write the given information. 

The sum of sequence:

$\sum _{k=8}^{40}(-3k)$

Step 2. Use the property of sequences.

Using the property of sequences:$$\begin{gathered} \underset{k=1}{\overset{n}{\sum \left(c{a}_k\right)}}=c\underset{k=1}{\overset{n}{\sum a_k}} \\ So, \\ \underset{k=1}{\overset{n}{\sum (-3k)}}=-3\underset{k=1}{\overset{n}{\sum k}} \end{gathered}$$

Step 3. Use the formula for sum of sequences of n real numbers 

$$\begin{gathered} \underset{k=8}{\overset{40}{\sum k}}=\underset{k=1}{\overset{40}{\sum k}}\underset{k=1}{\overset{7}{-\sum k}} \\ \Rightarrow \underset{k=8}{\overset{40}{\sum k}}=\frac{40(40+1)}{2}-\frac{7(7+1)}{2} \\ \Rightarrow \underset{k=8}{\overset{40}{\sum k}}=20\times 41-7\times 4 \\ \Rightarrow \underset{k=8}{\overset{40}{\sum k}}=820-28 \\ \Rightarrow \underset{k=8}{\overset{40}{\sum k}}=792 \end{gathered}$$

Step 4. Use the property of sequences from Step 2.

Using the property of sequences gives:

$$\begin{gathered} \underset{k=8}{\overset{40}{\sum k}}=792 \\ \Rightarrow (-\underset{k=8}{\overset{40}{3)\sum k}}=-3\times 792 \\ \Rightarrow \underset{k=8}{\overset{40}{(-3)\sum k}}=-2376 \end{gathered}$$