Chapter 12: Q 82. (page 809)

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

Short Answer

Expert verified

The sum of this sequence is 89964.

Step by step solution

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

Short Answer

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.∑k3k=424

Short Answer

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.

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Calculus

Statistics

Geometry

Logic and Functions

Decision Maths

Pure Maths

Study anywhere. Anytime. Across all devices.

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