Question

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

$\underset{k=5}{\overset{20}{\sum k^3}}$

Short answer

The sum of this sequence is 44000.

Step-by-step solution

Step 1. Write the given information.

The sum of sequence:

$\underset{k=5}{\overset{20}{\sum k^3}}$

Step 2. Use the properties of sequence.

Using the properties of sequence:

$$\begin{gathered} \underset{k=j+1}{\overset{n}{\sum k^3}}=\underset{k=1}{\overset{n}{\sum k^3}}\underset{k=1}{\overset{j}{-\sum k^3}} \\ So, \\ \underset{k=5}{\overset{20}{\sum k^3}}=\underset{k=1}{\overset{20}{\sum k^3}}\underset{k=1}{\overset{4}{-\sum k^3}} \end{gathered}$$

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,

$$\begin{gathered} \underset{k=5}{\overset{20}{\sum k^3}}=\underset{k=1}{\overset{20}{\sum k^3}}\underset{k=1}{\overset{4}{-\sum k^3}} \\ \Rightarrow \underset{k=5}{\overset{20}{\sum k^3}}={\left[\frac{20(20+1)}{2}\right]}^2-{\left[\frac{4(4+1)}{2}\right]}^2 \\ \Rightarrow \underset{k=5}{\overset{20}{\sum k^3}}={\left[10\times 21\right]}^2-{\left[2\times 5\right]}^2 \\ \Rightarrow \underset{k=5}{\overset{20}{\sum k^3}}=44100-100 \\ \Rightarrow \underset{k=5}{\overset{20}{\sum k^3}}=44000 \end{gathered}$$