Book edition 6th

Author(s) Sullivan

Pages 1200 pages

ISBN 9780321795465

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Short Answer

In Problems 71-82, find the sum of each sequence.
${\sum k^{3}}_{k = 4}^{24}$

The sum of this sequence is 89964.

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1. Write the given information.

The sum of sequence:

${\sum k^{3}}_{k = 4}^{24}$

Step 2. Use the properties of sequence.

Using the properties of sequence:

${\sum k^{3}}_{k = j + 1}^{n} = {\sum k^{3}}_{k = 1}^{n} {- \sum k^{3}}_{k = 1}^{j} S o, {\sum k^{3}}_{k = 4}^{24} = {\sum k^{3}}_{k = 1}^{24} {- \sum k^{3}}_{k = 1}^{3}$

Step 3. Use the formula for sum of sequences of (n3 ), n - real numbers.

The formula for summation: ${\sum k^{3}}_{k = 1}^{n} = 1^{3} + 2^{3} + . . . + n^{3} \Rightarrow {\sum k^{3}}_{k = 1}^{n} = {[\frac{n (n + 1)}{2}]}^{2}$

So,

${\sum k^{3}}_{k = 4}^{24} = {\sum k^{3}}_{k = 1}^{24} {- \sum k^{3}}_{k = 1}^{3} \Rightarrow {\sum k^{3}}_{k = 4}^{24} = {[\frac{24 (24 + 1)}{2}]}^{2} - {[\frac{3 (3 + 1)}{2}]}^{2} \Rightarrow {\sum k^{3}}_{k = 4}^{24} = {[12 \times 25]}^{2} - {[3 \times 2]}^{2} \Rightarrow {\sum k^{3}}_{k = 4}^{24} = 90000 - 36 \Rightarrow {\sum k^{3}}_{k = 5}^{20} = 89964$

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Precalculus Enhanced with Graphing Utilities

Answers without the blur.

Short Answer

In Problems 71-82, find the sum of each sequence.
${\sum k^{3}}_{k = 4}^{24}$

Step by Step Solution

Step 1. Write the given information.

Step 2. Use the properties of sequence.

Step 3. Use the formula for sum of sequences of (n3 ), n - real numbers.

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Precalculus Enhanced with Graphing Utilities

Answers without the blur.

Short Answer

In Problems 71-82, find the sum of each sequence.∑k3k=424

Step by Step Solution

Step 1. Write the given information.

Step 2. Use the properties of sequence.

Step 3. Use the formula for sum of sequences of (n3 ), n - real numbers.

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In Problems 71-82, find the sum of each sequence.
${\sum k^{3}}_{k = 4}^{24}$