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

[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

 

[cepheid]

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.

 

[HallsofIvy]

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).