Hi there--
I'll demonstrate the proof by induction method for #3. Then, you try the others by applying
the same method. Email me offline of you get stuck and I'll give you some more help.
The Problem:
3) Prove that n(n+1)(n+2) is an integer multiple of 6.
The Solution:
We will use the proof by induction method. You did not mention this in your problem
statement, but we assume that n is a positive integer, n = 1, 2, 3, and so on.
Think of an induction proof as climbing a staircase. The first step of any induction proof is to
show that the statement is true for a specific step on the staircase. We will show that the
statement is true when n = 1.
Let n = 1.
n(n+1)(n+2) = (1)(2)(3) = 6.
We know that 6*1=6, so 6 is an integer multiple of 6. The statement is true when n = 1.
Now we are going to show that when the statement is true for some value of n, that it is also
true for the next value of n. Let's have that "any value of n" be k. Then the next step after
n = k will be n = k+1.
This part of an induction proof is the one that is usually most confusing for folks.
We make the Induction Hypothesis: Suppose that n(n+1)(n+2) is an integer multiple of 6
when n=k.
We will show that it is an integer multiple of 6 when n = k+1.
This is not the same as the first part, where we named an actual number where the statement
is true. Now we say, "Let's assume that the statement is true, somewhere, out there in space,
we're not saying where, maybe I don't even know where; just somewhere."
If we can prove, assuming that n(n+1)(n+2) is an integer multiple of 6 for n = k, it is also an
integer multiple of 6 for n = k + 1 (that is, if it works on some step on the staircase, then it
must also work at the next step on the staircase), then, since we do know of a certain place
(n = 1) where the statement is true, we will have proved that it is true for all values of n.
OK, let's do it:
Assume that n(n+1)(n+2) is an integer multiple of 6 when n=k. In other words, assume
that k(k+1)(k+2) is an integer multiple of 6. We will show that n(n+1)(n+2) is also an integer
multiple of 6 when n=k+1. In other words, substitute k+1 for n and show that this expression
is an integer multiple of 6.
Now we are going to do a bunch of algebra to get this expression into a form that we can
show is an integer multiple of 6.
Multiply everything out using the distributive property.
Combine like terms.
This is where we use our Induction Hypothesis to help our proof. Notice that
. Multiply is out yourself to check. Rearrange our equation to
separate this polynomial.
By the Induction Hypothesis, 
is also an integer multiple of 6. Then we will have
an integer multiple of 6 plus an integer multiple of 6 which is also an integer multiple of 6.
Factor a 3 out of the second polynomial.
Now we have 3 * (k+1) * (k+2). Notice that (k+1) and (k+2) are consecutive integers. One of
them is even and one of them is odd. By definition, the even integer is an integer
multiple of 2. Thus we have three factors: 3, an integer multiple of 2 and an odd integer.
Their product must be an integer multiple of 6.
The expression (n)(n+1)(n+2) is an integer multiple of 6 when n = k+1. Therefore the
expression (n)(n+1)(n+2) is an integer multiple of 6 for all positive integers n.
That's it. Now you try the other problems using these steps:
1) Prove the statement is true for an initial value of n (usually n=0 or n=1)
2) Make the Induction Hypothesis: assume that the statement is true for n=k.
3) Show that the statement if n=k is true then n=k+1 is also true.
Feel free to email me at the address below if questions arise. Good luck!
Mrs. Figgy
math.in.the.vortex@gmail.com
