Hi. Need some guidance to make the trapezoid rule more effective.

Let f:[a,b]\to\mathbb{R} be a function. I want to find an approximation to the integral

\int_{a}^b f(x)dx.

I know that the trapezoid rule is given by

I_n =h\sum_{i=1}^{2^n} \frac{f(x_{i-1})+f(x_i)}{2} where h=\frac{b-a}{2^n}.
(Notice that I want 2^n trapezoids/parts)

I_n can be rewritten as

I_n =h[\frac{f(a)+f(b)}{2}+\sum_{i=1}^{2^n -1}f(a+ih)].

Ok, now over to the problem. To make the method more effective I use the approximation to integral from I_{n-1} in I_n in this way

I_n =\frac{I_{n-1}}{2}+h\sum_{i=1}^{2^{n-1}}f(a+(2i-1)h).

The part that I dont understand in this new formula is why you just can divide \frac{I_{n-1}}{2}? The other part I do understand, because in I_{n-1} you have already calculated f(x_i) where i is even and i\in[0,2^n], so in I_n you only need to calculate f(x_i) with odd indexes i.

I hope the problem is clear!? And I hope someone can help me explain and understand this formula. Maby it's a good idea to prove that it works? I dont know how, but maby a hint could help?