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Q 82.

Expert-verified
Found in: Page 809

### Precalculus Enhanced with Graphing Utilities

Book edition 6th
Author(s) Sullivan
Pages 1200 pages
ISBN 9780321795465

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

The sum of this sequence is 89964.

See the step by step solution

## Step 1. Write the given information.

The sum of sequence:

$\underset{k=4}{\overset{24}{\sum {k}^{3}}}$

## Step 2. Use the properties of sequence.

Using the properties of sequence:

$\underset{k=j+1}{\overset{n}{\sum {k}^{3}}}=\underset{k=1}{\overset{n}{\sum {k}^{3}}}\underset{k=1}{\overset{j}{-\sum {k}^{3}}}\phantom{\rule{0ex}{0ex}}So,\phantom{\rule{0ex}{0ex}}\underset{k=4}{\overset{24}{\sum {k}^{3}}}=\underset{k=1}{\overset{24}{\sum {k}^{3}}}\underset{k=1}{\overset{3}{-\sum {k}^{3}}}\phantom{\rule{0ex}{0ex}}$

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

The formula for summation:$\underset{k=1}{\overset{n}{\sum {k}^{3}}}={1}^{3}+{2}^{3}+...+{n}^{3}\phantom{\rule{0ex}{0ex}}⇒\underset{k=1}{\overset{n}{\sum {k}^{3}}}={\left[\frac{n\left(n+1\right)}{2}\right]}^{2}$

So,

$\underset{k=4}{\overset{24}{\sum {k}^{3}}}=\underset{k=1}{\overset{24}{\sum {k}^{3}}}\underset{k=1}{\overset{3}{-\sum {k}^{3}}}\phantom{\rule{0ex}{0ex}}⇒\underset{k=4}{\overset{24}{\sum {k}^{3}}}={\left[\frac{24\left(24+1\right)}{2}\right]}^{2}-{\left[\frac{3\left(3+1\right)}{2}\right]}^{2}\phantom{\rule{0ex}{0ex}}⇒\underset{k=4}{\overset{24}{\sum {k}^{3}}}={\left[12×25\right]}^{2}-{\left[3×2\right]}^{2}\phantom{\rule{0ex}{0ex}}⇒\underset{k=4}{\overset{24}{\sum {k}^{3}}}=90000-36\phantom{\rule{0ex}{0ex}}⇒\underset{k=5}{\overset{20}{\sum {k}^{3}}}=89964\phantom{\rule{0ex}{0ex}}$