For calculating divided (fraction) difference table for interpolating $(x_i, f_i)$, $i=1,2,...,n$; by using a polynomial with degree lower or equal to $n$, $n(n+1)/2$ difference fraction was used.
I think this sentence is false and $n(n-1)/2$ is needed. Am I right? any hint or idea would highly appreciated.
I wouldn't call it a "big contradiction". If you count the divided differences of orders $1,2,\dots n-1$, then indeed there are $n(n-1)/2$ of them. But if for some reason one includes the values $f_i$ themselves as "zero-order differences", then the total is $n(n+1)/2$.
Compare with derivatives: the derivative of order $0$ is the function itself. This convention is useful for writing down Taylor's formula. Something like that can be useful for divided differences too.