Question

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

$\sum _{k=1}^{40}k$

Short answer

The sum of this sequence is 820.

Step-by-step solution

Step 1. Write the given information.

The sum of sequence:$\sum _{k=1}^{40}k;k=1,...,40$

Step 2. Use the formula for sum of sequence of n real numbers. 

The formula for summation is:$$\begin{gathered} \underset{k=1}{\overset{n}{\sum k}}=1+2+...+n \\ \Rightarrow \underset{k=1}{\overset{n}{\sum k}}=\frac{n(n+1)}{2} \end{gathered}$$

Step 3. Use the formula with c=k and n=40.

Using the formula gives:

$$\begin{gathered} \underset{k=1}{\overset{40}{\sum k}}=\frac{40\times (40+1)}{2} \\ \Rightarrow \underset{k=1}{\overset{40}{\sum k}}=\frac{40\times \left(41\right)}{2} \\ \Rightarrow \underset{k=1}{\overset{40}{\sum k}}=\frac{1640}{2} \\ \Rightarrow \underset{k=1}{\overset{40}{\sum k}}=820 \end{gathered}$$