Hi Hossun.

We need to give this a name so let f(n) = 1/(1x2)+1/(2x3)+1/(3x4)...+1/(n(n+1)).

The first thing we notice is that for n > 1, we are just adding another fraction to the previous value of f(n). So we can construct f(n) = f(n-1) + 1/(n(n+1)).

Now look at the small values of n:
f(1) = 1/2,
f(2) = 1/2 + 1/6 = 2/3,
f(3) = 2/3 + 1/12 = 3/4,
f(4) = 3/4 + 1/20 = 4/5, etc.

So for the first few small values of n, we have proven by demonstration that f(n) = n / (n+1).

Our task is to prove that if it works for any positive integer value of n, then it works for n + 1. This way, it must by induction work for all subsequent values of n.

Formally said, we need to prove that if for some positive integer n we can show that f(n) = n / (n+1), then we can conclude that f(n+1) = (n + 1) / (n + 2).

We begin the real "proof" by expanding f(n + 1):

f(n + 1) = f(n) + 1 / ((n+1)((n+1)+1)) because that's based on the construction.

= n / (n+1) + 1 / ((n+1)(n+2)) because f(n) = n / (n+1); this is called "using what you know from earlier".

= n(n+2) / ((n+1)(n+2)) + 1 / ((n+1)(n+2)) because we can multiply the left fraction by (n+2)/(n+2).

= (n² + 2n + 1) / ((n+1)(n+2)) because we have a common denominator and can combine the numerators.

= (n+1)² / ( (n+1)(n+2)) because we can factor the numerator now; it is a perfect square.

= (n+1) / (n+2) because we can cancel the common (n+1) factor from the numerator and denominator.

Q.E.D. (which means "that which was to be proven", in other words: "voilà")

Cheers,
Stephen La Rocque.