Chapter 12: Q 58. (page 809)

In Problems 51–60, write out each sum.
$\sum_{k = 0}^{n - 1} (2 k + 1)$

Short Answer

Expert verified

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

Step by step solution

Step 1. Write the given information.

The given summation notation is:

$\sum_{k = 0}^{n - 1} (2 k + 1)$

Step 2. Write out the sum.

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

$\sum_{k = 0}^{n - 1} (2 k + 1) = \{(2 \cdot 0 + 1) + (2 \cdot 1 + 1) + (2 \cdot 2 + 1) + . . . . . . . + (2 \cdot (n - 1) + 1)\} = \{(0 + 1) + (2 + 1) + (4 + 1) + . . . . . . . + (2 n - 2 + 1)\} = \{1 + 3 + 5 + . . . . . . . + (2 n - 1)\}$

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

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In Problems 51–60, write out each sum.
$\sum_{k = 0}^{n - 1} (2 k + 1)$

Short Answer

Step by step solution

Step 1. Write the given information.

Step 2. Write out the sum.

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In Problems 51–60, write out each sum.∑k=0n-12k+1

Short Answer

Step by step solution

Step 1. Write the given information.

Step 2. Write out the sum.

One App. One Place for Learning.

Most popular questions from this chapter

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In Problems 51–60, write out each sum.
$\sum_{k = 0}^{n - 1} (2 k + 1)$