# Difference between revisions of "An Introduction to Mathematical Induction: The Sum of the First n Natural Numbers, Squares and Cubes."

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that | that | ||

− | :: <math style="vertical-align: 0px">{\displaystyle \frac{n(n+1)(2n+1)}{6}\,=\,\frac{1(1+1)(2+1)}{6}\,=\,1. | + | :: <math style="vertical-align: 0px">{\displaystyle \frac{n(n+1)(2n+1)}{6}\,=\,\frac{1(1+1)(2+1)}{6}\,=\,1,} </math> |

+ | |||

+ | so the rule is certainly true when <math style="vertical-align: -1px">n=1.</math> | ||

This gives us our starting point. For the induction step, let's assume | This gives us our starting point. For the induction step, let's assume |

## Revision as of 08:52, 22 August 2015

## Sigma Notation

In math, we frequently deal with large sums. For example, we can write

which is a bit tedious. Alternatively, we may use ellipses to write this as

However, there is an even more powerful shorthand known as **sigma notation**. When we write

this means the same thing as the previous two mathematical statements.
Here, the **index** below the capital sigma, , is the letter , and the that follows the is our **rule** to
apply to each value of . The values and tell us how many times to repeat the rule, i.e., to follow the rule for
then add the rule for then for and continue in this
manner until you reach . In other words,

Of course, we can change the rule and/or the index. For example,

Most importantly, we frequently don't have the luxury of bounds that are actual values. We can also write something like

or

These non-fixed indices allow us to find rules for evaluating some important sums.

## Proof by (Weak) Induction

When we count with **natural or counting numbers** (frequently denoted ), we begin with one, then keep adding one unit at a
time to get the next natural number. We then add one to that result to get the next natural number, and continue in this manner. In other words,

This is the basis for weak, or simple induction; we must first prove our conjecture is true for the lowest value (such as ), and then show whenever it's true for an arbitrary it's true for as well. This mimics our development of the natural numbers.

It is also equivalent to prove that whenever the conjecture is true for it's true for Which approach you choose can depend on which is more convenient, or frequently which is more appealing to the teacher grading the work.

Although we won't show examples here, there are induction proofs that require **strong induction**. This occurs when proving it for the case requires assuming more than just the
case. In such situations, strong induction assumes that the conjecture is true for ALL cases of lower value than down to our base case.

## The Sum of the first *n* Natural Numbers

* Claim.* The sum of the first natural numbers is

* Proof.* We must follow the guidelines shown for induction arguments.
Our base step is and plugging in we find that

Which is clearly the sum of the single integer . This gives us our starting point. For the induction step, let's assume the claim is true for so

Now, we have

as required.

## The Sum of the first *n* Squares

* Claim.* The sum of the first squares is

* Proof.* Again, our base step is and plugging in we find
that

so the rule is certainly true when

This gives us our starting point. For the induction step, let's assume the claim is true for so

Now, we have

as required.

## The Sum of the first *n* Cubes

* Claim.* The sum of the first cubes is

Notice that the formula is really similar to that for the first natural numbers.

* Proof.* Plugging in we find that

completing our base step.

For the induction step, let's assume the claim is true for so

Now, we have

as required.

Aside from being good examples of proof by simple or weak induction, these formulas are useful to find an integral as a limit of a Riemann sum.