# Proving Mathematical Induction: Solving Equations with Step-by-Step Guide

• sierra.arreis
That will make it clear that (n+1)(n+2) is a factor of both terms, and you can factor it out, making the terms simpler. That's a common technique in algebra, to simplify before you try to prove something. However, it's not clear to me what you're trying to prove, because your original formula isn't there. If you're trying to prove that \sum_{i=1}^n i(i+1)= n(n+1)(n+2)/3by mathematical induction, then you need to prove that if it is true for some value of n, it is true for n+1. That's what I was doing above.

#### sierra.arreis

The equation is: [i(i+1)=n(n+1)(n+2)/3] whereas i=1

so the beginning process would be 2+6+12+20...+n(n+1)=n(n+1)(n+2)/3

after the equation is proven for n=1 [(1(1+1)=1(1+1)(1+2)/3] then we must prove for n=n+1

thats where i begin to stop understanding.
So...
2+6+12+20...+n(n+1)+n+1(n+1+1)??= n(n+1)(n+2)/3+n+1??

if my nonsense is correct then we are attempting to prove that n(n+1)(n+2)/3+n+1=(n+1)(n+2)

...help

It seems like you're trying to prove that:

$$\sum_{i=1}^n i(i+1) = \frac{n(n+1)(n+2)}{3}$$

Every inductive proof has the same form. You start with a base case. In this case, n = 1.

Is it true that 1(2) = 1(2)(3)/3? Yes, so it works for the base case.

Then the argument is that if you can show that **IF** it works for n, **THEN** it works for n+1, then it must be true for all n by mathematical induction.

In other words, ASSUMING that it works for n, show that it works for n+1. If you can do that, then you can state that by mathematical induction, it works for all n.

sierra.arreis said:
The equation is: [i(i+1)=n(n+1)(n+2)/3] whereas i=1
So you mean
$$\sum_{i= 1}^n i(i+1)= n(n+1)(n+2)/3$$

so the beginning process would be 2+6+12+20...+n(n+1)=n(n+1)(n+2)/3

after the equation is proven for n=1 [(1(1+1)=1(1+1)(1+2)/3] then we must prove for n=n+1
Well, not "n= n+1", that's meaning less. You mean to prove that if it is true for a given n, it is true for n+1.

thats where i begin to stop understanding.
So...
2+6+12+20...+n(n+1)+n+1(n+1+1)??= n(n+1)(n+2)/3+n+1??

if my nonsense is correct then we are attempting to prove that n(n+1)(n+2)/3+n+1=(n+1)(n+2)

...help
$$\sum_{i=1}^{n+1} i(i+1)= \sum_{i=1}^n i(i+1)+ (n+1)(n+1+1)= n(n+1)(n+2)/3+ (n+1)(n+2)$$
and you want to prove that is equal to (n+1)(n+2)(n+3)/3, the basic formula n(n+1)(n+2)/3 with "n" replaced by "n+1".

I would recommend that the first thing you do is factor (n+1) and (n+2) out of n(n+1)(n+2)/3+ (n+1)(n+2).

## What is mathematical induction?

Mathematical induction is a method used to prove that a statement or property holds for all natural numbers. It involves proving a base case and then using the fact that the statement holds for a particular number to show that it also holds for the next number.

## How does mathematical induction work?

Mathematical induction works by proving a statement for a base case, typically when the input is 0 or 1. Then, we assume that the statement holds for a particular natural number n, and use this assumption to show that it also holds for the next number n+1. This process is repeated until we can show that the statement holds for all natural numbers.

## What is the difference between strong and weak induction?

Strong induction is a variation of mathematical induction where instead of just assuming that the statement holds for a particular natural number, we assume that it holds for all natural numbers up to that number. This allows us to make a stronger inductive step and can be useful for proving more complex statements. Weak induction, on the other hand, only assumes that the statement holds for a particular number and the next number, without making any assumptions about all numbers in between.

## What are some common mistakes when using mathematical induction?

Some common mistakes when using mathematical induction include not proving the base case, using the wrong inductive hypothesis, and incorrectly applying the inductive step. It is important to carefully follow the steps of mathematical induction and double check all assumptions and conclusions to avoid these mistakes.

## What types of statements can be proved using mathematical induction?

Mathematical induction is commonly used to prove statements about natural numbers, such as number patterns, divisibility, and algebraic equations. It can also be used to prove statements about other mathematical concepts, such as graphs, sets, and functions, as long as they can be translated into statements about natural numbers.