1. Mathematical Induction

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

Could someone please show me how to do the intermediate steps:
Assume the results is true for n = k, where k is some positive integer.
i.e.

$\frac{1}{1*2*3} + \frac{1}{2*3*4} + ... + \frac{1}{k(k+1)(k+2)} = \frac{k(k+3)}{4(k+1)(k+2)}$
R.T.P.
$\frac{1}{1*2*3} + \frac{1}{2*3*4} + ... + \frac{1}{(k+1)(k+2)(k+3)} = \frac{(k+1)(k+4)}{4(k+2)(k+3)}$
But $\frac{1}{1*2*3} + \frac{1}{2*3*4} + ... + \frac{1}{(k+1)(k+2)(k+3)}$
= $\frac{k(k+3)}{4(k+1)(k+2)}+\frac{1}{(k+1)(k+2)(k+3 )}$
= $\frac{k(k+3)^2}{4(k+1)(k+2)(k+3)}$

...

I cant go any further!

2. Originally Posted by Sunyata
$
\frac{1}{1*2*3} + \frac{1}{2*3*4} + ... + \frac{1}{n(n+1)(n+2)} = \frac{n(n+3)}{4(n+1)(n+2)}
$

Could someone please show me how to do the intermediate steps:
Assume the results is true for n = k, where k is some positive integer.
i.e.

$\frac{1}{1*2*3} + \frac{1}{2*3*4} + ... + \frac{1}{k(k+1)(k+2)} = \frac{k(k+3)}{4(k+1)(k+2)}$
R.T.P.
$\frac{1}{1*2*3} + \frac{1}{2*3*4} + ... + \frac{1}{(k+1)(k+2)(k+3)} = \frac{(k+1)(k+4)}{4(k+2)(k+3)}$
But $\frac{1}{1*2*3} + \frac{1}{2*3*4} + ... + \frac{1}{(k+1)(k+2)(k+3)}$
= $\frac{k(k+3)}{4(k+1)(k+2)}+\frac{1}{(k+1)(k+2)(k+3 )}$
= $\frac{k(k+3)^2}{4(k+1)(k+2)(k+3)}$ $\leftarrow$ should be $\frac{k(k+3)^2 + 4}{4(k+1)(k+2)(k+3)}$

...

I cant go any further!
$= \frac{k(k+3)^2 + 4}{4(k+1)(k+2)(k+3)}$

$= \frac{k^3 + 6k^2 + 9k + 4}{4(k+1)(k+2)(k+3)}$

$= \frac{k^3 + 4k^2 + 2k^2 + 8k + k + 4}{4(k+1)(k+2)(k+3)}$

$= \frac{k(k^2 + 2k + 1) +4(k^2 + 2k + 1)}{4(k+1)(k+2)(k+3)}$

$= \frac{(k^2 + 2k + 1)(k + 4)}{4(k+1)(k+2)(k+3)}$

$= \frac{(k + 1)^2(k + 4)}{4(k+1)(k+2)(k+3)}$

$= \frac{(k + 1)(k + 4)}{4(k+2)(k+3)}$

3. Originally Posted by Sunyata
$
\frac{1}{1*2*3} + \frac{1}{2*3*4} + ... + \frac{1}{n(n+1)(n+2)} = \frac{n(n+3)}{4(n+1)(n+2)}
$

Could someone please show me how to do the intermediate steps:
Assume the results is true for n = k, where k is some positive integer.
i.e.

$\frac{1}{1*2*3} + \frac{1}{2*3*4} + ... + \frac{1}{k(k+1)(k+2)} = \frac{k(k+3)}{4(k+1)(k+2)}$
R.T.P.
$\frac{1}{1*2*3} + \frac{1}{2*3*4} + ... + \frac{1}{(k+1)(k+2)(k+3)} = \frac{(k+1)(k+4)}{4(k+2)(k+3)}$
But $\frac{1}{1*2*3} + \frac{1}{2*3*4} + ... + \frac{1}{(k+1)(k+2)(k+3)}$
= $\frac{k(k+3)}{4(k+1)(k+2)}+\frac{1}{(k+1)(k+2)(k+3 )}$
= $\frac{k(k+3)^2}{4(k+1)(k+2)(k+3)}$

...

I cant go any further!
= $\frac{k(k+3)}{4(k+1)(k+2)}+\frac{1}{(k+1)(k+2)(k+3 )}$
= $\frac{k(k+3)^2}{4(k+1)(k+2)(k+3)}+\frac{4}{4(k+1)( k+2)(k+3)}$
= $\frac{k^3+6k^2+9k+4}{4(k+1)(k+2)(k+3)}$
= $\frac{(k+1)(k+1)(k+4)}{4(k+1)(k+2)(k+3)}$