Question

In Problems 51–60, write out each sum.

$\sum _{k=0}^{n-1}\left(2k+1\right)$

Short answer

The required sum is$\sum _{k=0}^{n-1}\left(2k+1\right)=\left\{1+3+5+.......+\left(2n-1\right)\right\}$.

Step-by-step solution

Step 1. Write the given information.

The given summation notation is:

$\sum _{k=0}^{n-1}\left(2k+1\right)$

Step 2. Write out the sum.

To get the sum, we need to substitute $0,1,2,.......,\left(n-1\right)$for k,in the given sequence $\sum _{k=0}^{n-1}\left(2k+1\right)$,

$$\begin{gathered} \sum _{k=0}^{n-1}\left(2k+1\right)=\left\{\left(2·0+1\right)+\left(2·1+1\right)+\left(2·2+1\right)+.......+\left(2·\left(n-1\right)+1\right)\right\} \\ =\left\{\left(0+1\right)+\left(2+1\right)+\left(4+1\right)+.......+\left(2n-2+1\right)\right\} \\ =\left\{1+3+5+.......+\left(2n-1\right)\right\} \end{gathered}$$

Therefore,$\sum _{k=0}^{n-1}\left(2k+1\right)=\left\{1+3+5+.......+\left(2n-1\right)\right\}$.