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BITSAT Exam - Math: Study Guide & Test Prep32 chapters | 299 lessons | 18 flashcard sets

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Lesson Transcript

Instructor:
*David Karsner*

If you wanted to know the sum of the cubes of the first n natural numbers, you could find all the cubes and then add them all together. After the first few cubes, the numbers get large and time- consuming. There are shortcuts to finding this sum.

On many occasions in mathematics, you'll probably catch yourself detecting patterns. These patterns, when applied to a repetitive process, can allow you to complete the computations quickly. Finding the sum of the cubes of the first *n* natural numbers is one of those occasions. This lesson will show the pattern that will allow you to find the sum of the first *n* natural numbers quickly. It'll also give a brief proof that the pattern does work in all cases.

If you've been around algebra for any length of time, you've probably heard the phrase **Sum of Cubes** when referring to the factoring of two cubed terms that have been added together. *x*3 + 27 would be an example of this kind of sum of cubes. That is not what this lesson is about. This lesson is about finding the sum of the cubes of the first *n* natural numbers. If you wanted to find the sum of the first three (*n* = 3) cubes, it would be (1)3 + 23 + 33 or 1 + 8 + 27 = 36. Since *n* was only 3, it didn't take very long to figure out that the sum of the first three cubes was 36. When *n* gets larger it becomes a lot more time consuming to find the sum. There's a formula that allows for the sum of the first *n* cubes to be tallied without a lot of time or trouble. As you can see, the formula is read as:

Let's start with *n* = 3, since we already know the answer.

(3(3+1)/2)2

((3 x 4)/2)2

(12/2)2

(6)2

Therefore, our answer is 36.

Find the sum of the first 12 cubes.

*n* = 12

*n* + 1 = 13

12 x 13 = 156

156 / 2 = 78

782 = 6,084

That's a lot quicker than finding the 12 cubes and then adding them together, and you get the same answer as doing it longhand.

There's another means of finding the sum of the cubes of the first *n* natural numbers. You can sum up the first *n* numbers and then square your answer. This may sound simpler at first, but summing up those first *n* numbers can be very time-consuming. If you let *n*=12, you would need to add 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 + 11 + 12 = 78 and then square it; 782 = 6,084. The other way is quicker.

Induction can be used to prove that the sum of the first *n* natural numbers is the square of:

((*n* x (*n*+1)) / 2)2

The first step of induction is to prove that when *n* = 1, it'll work. So *n* = 1, *n* + 1 = 2. (1)(2)/(2) = 1. The square of 1 = 1. 13 = 1. It works for *n* = 1.

In the second step of induction, you assume that the theory works for *n*. So, we assume that ((*n*(*n* + 1))/2)2 = the sum of the first *n* natural cubes.

The third part of induction is the one that takes some algebraic manipulations. In part two, we assume that the theory is correct; in part three, we have to show that adding our next cube to our assumption will gives us our assumption with *n* + 1 instead of *n*. The image on your screen right now will show you how to set up step three of the proof by induction. The remainder of algebraic manipulations used to determine that step three is correct and that the sum of the first *n* cubes is *n* x (*n* + 1) / 2 is not provided here.

Through some moderately heavy algebraic manipulations, we end up with (*x* + 3)3 on the left side of the equation that equals (*x* + 3)3 on the right side of the equation. This proves step three and therefore the original theory. Simple, right?

Let's take a few moments to review what we've learned about the sum of the cubes of the first *n* natural numbers. First, recall that the phrase **Sum of Cubes** is used when referring to the factoring of two cubed terms that have been added together. To find the sum of the cubes of the first *n* natural numbers, you could find the cubes of each number and then add all of them together. You could do it this way, or you could save yourself a lot of time and effort by using these steps:

- Take your number
*n*and multiply it by the next natural number.*n*(*n*+ 1). - Divide your answer from step one by 2.
- Square your answer from step 2.

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BITSAT Exam - Math: Study Guide & Test Prep32 chapters | 299 lessons | 18 flashcard sets

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