Mathematical induction help

$\displaystyle \frac{1}{1*2} + \frac{1}{2*3} + \frac{1}{3*4} + ... + \frac{1}{(n(n+1))} = \frac{n}{n+1}$

I have evaluated it for 1:

$\displaystyle \frac{1}{(1(1+1))} = \frac{1}{1+1} = \frac{1}{2}$

same with k:

$\displaystyle \frac{1}{(k(k+1))} = \frac{k}{k+1}$

now with k+1:

$\displaystyle \frac{1}{(k+1)(k+1+1)} = \frac{k+1}{k+1+1}$

$\displaystyle \frac{1}{(k+1)(k+2)} = \frac{k+1}{k+2}$

I don't really understand induction, I've looked up examples on google and such and I can't grasp it at all. I understand kind of how it works, but not how to do it.

A quick point in the right direction would be appreciated :)

Re: Mathematical induction help

Quote:

Originally Posted by

**bnosam** $\displaystyle \frac{1}{1*2} + \frac{1}{2*3} + \frac{1}{3*4} + ... + \frac{1}{(n(n+1))} = \frac{n}{n+1}$

I have evaluated it for 1:

$\displaystyle \frac{1}{(1(1+1))} = \frac{1}{1+1} = \frac{1}{2}$

The above is simply **the base case.**

For the inductive step suppose that for $\displaystyle K>1$ it is true that $\displaystyle \sum\limits_{n = 1}^K {\frac{1}{{n(n + 1)}}} = \frac{K}{{K + 1}}$

Now you use **that fact** to show that $\displaystyle \sum\limits_{n = 1}^{K+1} {\frac{1}{{n(n + 1)}}} = \frac{K+1}{{K + 1+1}}$ is also true.

Re: Mathematical induction help

Quote:

Originally Posted by

**bnosam** $\displaystyle \frac{1}{1*2} + \frac{1}{2*3} + \frac{1}{3*4} + ... + \frac{1}{(n(n+1))} = \frac{n}{n+1}$

I have evaluated it for 1:

$\displaystyle \frac{1}{(1(1+1))} = \frac{1}{1+1} = \frac{1}{2}$

same with k:

$\displaystyle \frac{1}{(k(k+1))} = \frac{k}{k+1}$

No, the left side is NOT a single fraction, it is a sum of fractions.

Quote:

now with k+1:

$\displaystyle \frac{1}{(k+1)(k+1+1)} = \frac{k+1}{k+1+1}$

The point is that the left side is just the previous sum with one added term:

since the previous sum was equal to k/(k+1), the new sum is k/(k+ 1)+ 1/[(k+1)(k+2)].

What do you get when you add those?

Quote:

$\displaystyle \frac{1}{(k+1)(k+2)} = \frac{k+1}{k+2}$

I don't really understand induction, I've looked up examples on google and such and I can't grasp it at all. I understand kind of how it works, but not how to do it.

A quick point in the right direction would be appreciated :)

Re: Mathematical induction help

Quote:

Originally Posted by

**bnosam** I don't really understand induction, I've looked up examples on google and such and I can't grasp it at all. I understand kind of how it works, but not how to do it.

Induction works by proving that a statement or equation is true for an infinite number of possible values that can be substituted into it. It works like a row of dominoes. Say you had set up a row of dominoes that theoretically was infinitely long, then you know that any domino will push the next one over, as long as the first domino is pushed. Mathematical induction works the same way - you prove a base case (which is equivalent to pushing the first domino) and show that IF the statement is true for an arbitrary value, THEN the statement will be true for the next value (which is equivalent to showing that any domino will push the next one over).

Re: Mathematical induction help

Quote:

Originally Posted by

**HallsofIvy** No, the left side is NOT a single fraction, it is a sum of fractions.

The point is that the left side is just the previous sum with one added term:

since the previous sum was equal to k/(k+1), the new sum is k/(k+ 1)+ 1/[(k+1)(k+2)].

What do you get when you add those?

Do you mean like t

$\displaystyle \frac{k}{k+1} + \frac{k+1}{k+1+1}$

$\displaystyle \frac{k}{k+1} + \frac{k+1}{k+2}$

$\displaystyle \frac{k(k+2) + (k+1)(k+1)}{(k+1)(k+2)}$

$\displaystyle \frac{k^2 + 2k + k^2 + 2k +1}{k^2 +3k + 2}$

$\displaystyle \frac{2k^2 + 4k + 1}{k^2 + 3k +2}$

Like that?

Re: Mathematical induction help

Quote:

Originally Posted by

**bnosam** Do you mean like t

$\displaystyle \frac{k}{k+1} + \frac{k+1}{k+1+1}$

Like that?

NO!

$\displaystyle \sum\limits_{n - = 1}^{K + 1} {\frac{1}{{n(n + 1)}}} =\\ \sum\limits_{n - = 1}^K {\frac{1}{{n(n + 1)}}} + \frac{1}{{(K + 1)(K + 2)}} =\\ \frac{K}{{K + 1}} + \frac{1}{{(K + 1)(K + 2)}} =\\ \frac{{K(K + 2) + 1}}{{(K + 1)(K + 2)}} = ?$

Re: Mathematical induction help

Quote:

Originally Posted by

**Plato** NO!

$\displaystyle \sum\limits_{n - = 1}^{K + 1} {\frac{1}{{n(n + 1)}}} =\\ \sum\limits_{n - = 1}^K {\frac{1}{{n(n + 1)}}} + \frac{1}{{(K + 1)(K + 2)}} =\\ \frac{K}{{K + 1}} + \frac{1}{{(K + 1)(K + 2)}} =\\ \frac{{K(K + 2) + 1}}{{(K + 1)(K + 2)}} = ?$

Ok I'm confused now as to where those come from.

But...

that bottom part is equal to:

$\displaystyle \frac{(k+1)(k+1)}{(k+1)(k+2}$

$\displaystyle = \frac{k+1}{k+2} $

Right?

Re: Mathematical induction help

Quote:

Originally Posted by

**bnosam** Ok I'm confused now as to where those come from.

But...

that bottom part is equal to:

$\displaystyle \frac{(k+1)(k+1)}{(k+1)(k+2}$

$\displaystyle = \frac{k+1}{k+2} $

Right?

Yes that is right and it is what you want to show.

NOTE THAT $\displaystyle \frac{k+1}{k+2} = \frac{k+1}{(k+1)+1} $