I know absolutely nothing about Brownian motion ( I wish I did) but I looked at Wikipedia, it made some sense how they explain it.

Heir.

It looks like a variation of the Riemann integral to me.

Let me give an example.

Say we want to find the quadradic variation of f(t)=t on the interval [0,1]

So we partition the region (like in the Riemann integral), I am going to be using left-endpoints because that is what they (Wikipedia) use.

Thus, divide the region into "n" parts. Each part has 1/n width.

The initial value is

x_0=0

x_1=0+1/n

x_2=0+1/n+1/n

x_3=0+1/n+1/n+1/n

...

You get the idea.

Thus,

x_k = k/n

Where 0<=k<=n-1

(Because of left-endpoints).

Now,

[f(x_1)-f(x_0)]^2=(1/n-0/n)^2=1/n^2

[f(x_2)-f(x_1)]^2=(2/n-1/n)^2=1/n^2

[f(x_3)-f(x_2)]^2=(3/n-2/n)^2=1/n^2

...

[f(x_{n-1}-f_{n-2}]^2=[(n-1)/n-(n-2)/n]^2=1/n^2

Now, we sum them up,

<f>_1 = SUM (of that whole thing) = 1/n^2+1/n^2+...+1/n^2=(n-1)/n^2

Now, we take the limit as n --> infinity.

This clearly goes to zero.

Thus, the quadradic variation of f(t)=t of 1 (meaning on interval [0,1]) is zero.

Now, they mention an interesting theorem.

If f(t) is differenciable, then its quadradic variation is zero.

Precisely what we got, zero.

I hope this helps somehow.