# Help on Mathematical Induction

1. Oct 11, 2006

### L²Cc

Proposition: 1*2*3+2*3*4+3*4*5+...+n(n+1)(n+2) = [n(n+1)(n+2)(n+3)]/4

Step (1): If n=1 then LHS (left hand side) = 6, and RHS = 6
Thus, P1 is true.

Step (2): If Pk is true then
k(k+1)(k+2) = [k(k+1)(k+2)(k+3)]/4

Now,
k(k+1)(k+2) + [k+1]([k+1]+1)([k+1]+2) = [k(k+1)(k+2)(k+3)]/4 + [k+1]([k+1]+1)([k+1]+2)

k(k+1)(k+2) +[k+1]([k+1]+1)([k+1]+2) = [k(k+1)(k+2)(k+3)]/4 + (k+1)(k+2)(k+3)

[k(k+1)(k+2)(k+3) + 4(k+1)(k+2)(k+3)]/4 ---> Common denominator - 4.

Then,
let k+1= A, k+2= B, k+3= C
k(A)(B)(C) + 4(A)(B)(C)/4
A(K+4)B(K+4)C(K+4)
Then,
(K+4)(ABC)
am i on the right track?
Then,
(k+4)(k+1)(k+2)(k+3)/4
Then?????? How do i prove that P(k+1) is true whenever Pk and P1 are true???!

2. Oct 11, 2006

### OlderDan

Are you sure about step 2?

3. Oct 11, 2006

### L²Cc

hmmm...yeah i think so...is there anything wrong with step two, other than i shouldve put in the sequence before the general nth term formula???!

4. Oct 11, 2006

### OlderDan

It is the sequence of products before the general term that was missing. It is also missing in the next step. Maybe you have kept track of it somehow, but I don't see it.

My guess is that to finish you need to expand
[n(n+1)(n+2)(n+3)]/4
with n=k+1 to look like
[k(k+1)(k+2)(k+3)]/4 + R
The assumption of truth for Pk tells you that you can replace [k(k+1)(k+2)(k+3)]/4 with the sum up through the k term. Is R what you need it to be for the next term to satisfy equality?

I can't stay around to work it out, but that is what I would try.

5. Oct 11, 2006

### L²Cc

Im confused...what's R in this equation?
i have added k+1 to the equation because im proving here that the proposition is applicable for the K+1th term too...(if you see what i mean?)

Is there a better way out other than expanding? (that would take for ages...)
Anyhow, you might not be reading this msg today........

6. Oct 11, 2006

### OlderDan

R is whatever is left over after you have isolated the first product. I had to drive to the airport when I left earlier. On the way I did the needed expansion in my head. Once you see it, you can too. You only have to distribute one of the binomials in the product; the rest can stay as binomials. You were really close to it in your earlier post. Do what I outlined, replacing n with k+1, and it will almost jump out at you if you don't ignore the early terms in the series.

7. Oct 12, 2006

### feod2003

isnt this java, if so use j creator

8. Oct 13, 2006

### L²Cc

you know what i just realized........i did prove that Pn is true:
(k+4)(k+1)(k+2)(k+3)/4 = ([k+1])([k+1]+1)([k+1]+2)([k+1]+3)/1=4, where k+1=n!
yey:

9. Oct 14, 2006

### OlderDan

This is not true. You are, and have been close to a proof, but this incorrect equation is not going to do it for you.

You seem to have used Pn to refer to the propositon. I suggest you use Sn to refer to the sum of terms in the series. So the proposition Pn is that
Pn: Sn = [n(n+1)(n+2)(n+3)]/4 where Sn = 1*2*3+2*3*4+3*4*5+...+n(n+1)(n+2)

You have correctly shown that P1 is true by showing that
P1: S1 = 1*2*3 = [1*2*3*4]/4 = 6

You will assume that Pk is true
Pk: Sk = [k(k+1)(k+2)(k+3)]/4

What you need to show is that given P1 is true and Pk is assumed true then P(k+1) is also true

P(k+1): S(k+1) = [(k+1)(k+2)(k+3)(k+4)]/4

The proof is completed by verifying this equality. To do that you need to write the left hand side and right hand side in terms of things that you know (or can assume) are equal. You need to write S(k+1) in terms of Sk and other terms. Then you need to manipulate the right side to identify something that is equal to Sk. If you then subtract Sk and its equivalent from both sides of the equation you will be comparing two expressions. If they are indeed equal, then you have verified the equality. If they are unequal, then you have disproved the proposition.

Last edited: Oct 14, 2006