Question

In Problems 71-82, find the sum of each sequence.

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

Short answer

The sum of this sequence is 89964.

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:

$$\begin{gathered} \underset{k=j+1}{\overset{n}{\sum k^3}}=\underset{k=1}{\overset{n}{\sum k^3}}\underset{k=1}{\overset{j}{-\sum k^3}} \\ So, \\ \underset{k=4}{\overset{24}{\sum k^3}}=\underset{k=1}{\overset{24}{\sum k^3}}\underset{k=1}{\overset{3}{-\sum k^3}} \end{gathered}$$

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

The formula for summation:
$$\begin{gathered} \underset{k=1}{\overset{n}{\sum k^3}}=1^3+2^3+...+n^3 \\ \Rightarrow \underset{k=1}{\overset{n}{\sum k^3}}={\left[\frac{n(n+1)}{2}\right]}^2 \end{gathered}$$

So,

$$\begin{gathered} \underset{k=4}{\overset{24}{\sum k^3}}=\underset{k=1}{\overset{24}{\sum k^3}}\underset{k=1}{\overset{3}{-\sum k^3}} \\ \Rightarrow \underset{k=4}{\overset{24}{\sum k^3}}={\left[\frac{24(24+1)}{2}\right]}^2-{\left[\frac{3(3+1)}{2}\right]}^2 \\ \Rightarrow \underset{k=4}{\overset{24}{\sum k^3}}={\left[12\times 25\right]}^2-{\left[3\times 2\right]}^2 \\ \Rightarrow \underset{k=4}{\overset{24}{\sum k^3}}=90000-36 \\ \Rightarrow \underset{k=5}{\overset{20}{\sum k^3}}=89964 \end{gathered}$$