Problem 56 - Precalculus with Limits

Question

In Exercises 49 - 58, find the sum using the formulas for the sums of powers of integers. \( \sum_{n=1}^{20}\left(n^3 - n\right) \)

Step-by-Step Solution

Answer

The value of the sum \( \sum_{n=1}^{20}\left(n^3 - n\right) \) is obtained after subtracting the calculated values of step 3 from step 2. Input the results into the final expression and calculate to get the final answer.

Step 1: Break apart the sum

Due to the properties of sums, the given sum \( \sum_{n=1}^{20}\left(n^3 - n\right) \) can be separated into two parts: \( \sum_{n=1}^{20} n^3 \) and \(- \sum_{n=1}^{20} n\)

Step 2: Use formula for sum of cubes

The sum \( \sum_{n=1}^{20} n^3 \) measures the sum of the cubes of the first 20 natural numbers. The formula for the sum of the cubes of the first 'n' natural numbers is \(\left(\frac{n(n+1)}{2}\right)^2\). Now plug \(n=20\) into the formula to get the sum of the cubes of the first 20 natural numbers.

Step 3: Use formula for sum of first 'n' natural numbers

The second part \(- \sum_{n=1}^{20} n \) is the sum of the first 20 natural numbers, but it is subtracted from the first part. The formula for the sum of the first 'n' natural numbers is \(\frac{n(n+1)}{2}\). Applying this formula with \(n=20\), we get the sum of the first 20 natural numbers, which will then be subtracted.

Step 4: Combine results

Now that results from Step 2 and Step 3 are obtained, subtract the sum of the first 20 natural numbers (found in step 3) from the sum of the cubes of the first 20 natural numbers (found in step 2).

Key Concepts

Sum of Cubes

The sum of cubes formula is a powerful tool to quickly calculate the sum of the cubes of the first 'n' natural numbers. When we talk about the "sum of cubes," we mean adding up numbers like 1 cubed, 2 cubed, all the way up to 'n' cubed. There's a neat formula for this, which is \[ \left(\frac{n(n+1)}{2}\right)^2 \]This formula reveals a beautiful pattern: the sum of cubes up to 'n' is actually the square of the sum of the first 'n' numbers! For example, when you cube each of the first 20 numbers and add them up, using this formula lets you do it quickly and accurately, without needing to cube and add each number manually.Understanding and applying this formula helps tackle sum-related problems efficiently, showing how mathematics can surprise us with elegant solutions.

Sum of First n Natural Numbers

Calculating the sum of the first 'n' natural numbers is an essential math skill. These numbers start from 1 and go up to 'n'. There's a quick formula for adding them all up:\[ \frac{n(n+1)}{2} \]This formula breaks down the problem of summing up these numbers into a very simple operation, saving time and reducing error. For instance, if you needed the sum of the numbers from 1 to 20, instead of adding each number together manually, you would just plug 20 into the formula, that is: \[ \frac{20 \times 21}{2} = 210 \] It is very handy in exercises like the one in the original problem where we need a fast way to compute the sum to solve more complex equations efficiently.

Properties of Sums

The properties of sums allow us to break down complicated expressions into simpler parts. This makes problem-solving much more manageable. One of the main properties is how we can split a sum into separate parts. For example:\[ \sum_{n=1}^{20} (n^3 - n) = \sum_{n=1}^{20} n^3 - \sum_{n=1}^{20} n \] This property of sums is critical as it lets us handle each part individually. After splitting, we can apply specific formulas to each part, solve them, and then combine the results. Such a property is especially useful in larger math problems where dealing with everything at once would be too complex.This approach helps with clarity and precision in calculations, illustrating how mathematics leverages its inherent properties for problem-solving.