Question

In Problems 61–70, express each sum using summation notation.

$1+3+5+7+....+\left[2\left(12\right)-1\right]$

Short answer

The required summation notation is $1+3+5+7+....+\left[2\left(12\right)-1\right]=\sum _{k=1}^n\left(2k-1\right)$.

Step-by-step solution

Step 1. Write the given information.

The given sum is:

$1+3+5+7+....+\left[2\left(12\right)-1\right]$

Step 2. Write the sum using summation notation.

Write each term in the same format:

$$\begin{gathered} 1+3+5+7+....+\left[2\left(12\right)-1\right]=\left[2\left(1\right)-1\right]+\left[2\left(2\right)-1\right]+\left[2\left(3\right)-1\right]+.....+\left[2\left(12\right)-1\right] \\ =\sum _{k=1}^n\left(2k-1\right) \end{gathered}$$

As we can see that the sum$\left[2\left(1\right)-1\right]+\left[2\left(2\right)-1\right]+\left[2\left(3\right)-1\right]+.....+\left[2\left(12\right)-1\right]$ has 12 terms, each term represents the form $2k-1$, and starts at $k=1$and ends at $k=12$.

Therefore,role="math" localid="1646761047379" $1+3+5+7+....+\left[2\left(12\right)-1\right]=\sum _{k=1}^n\left(2k-1\right)$