No, I'm sure it works out to be the same as long as you take the first interval to be y1 instead of y0.
For example, if the intervals were 1, 2, 3, 4, 5 and 6:
(y0, y1, y2, y3, y4, y5) Using h/3[ F + L + 2(even) + 4 (odd)]
= h/3 [y0 + y5 + 2(y2 + y4) + 4(y1 + y3)]
=h/3 [1 + 6 + 2(3 + 5)...