Simplify the sum $\sum_{i=1}^n(\frac{h_i}{2}(f(x_{i-1})+f(x_i)) +\frac{h_i^2}{12}((f'(x_{i-1})-f'(x_i)) $ $x_i$ equidistant

Question

Consider the interval $[a,b]$ with the nodes $a = x_0 < x_1 < \cdots < x_n = b$. Let $f \in C^1([a,b]) $ be a function.

Consider $$ Q_{[a,b]}(f) = \sum_{i=1}^n \left(\frac{h_i}{2}(f(x_{i-1})+f(x_i)\right) +\frac{h_i^2}{12}((f'(x_{i-1})-f'(x_i)) $$ with $h_i = x_i - x_{i-1} $.

Remark : ( $Q_{[a,b]}(f)$ is quadrature formula for $\int_a^b f(x)dx $ )

To do: Simplify $Q_{[a,b]}(f)$ for equidistant nodes $x_i$.

My work so far :

We have equidistant nodes $x_i$. So $x_i = a + i\frac{(b-a)}{n}$. We get that $h_i = \frac{(b-a)}{n}$ for all $i \in \{1,\ldots,n\}$. So:

\begin{align} & Q_{[a,b]}(f) = \sum_{i=1}^n\left(\frac{(\frac{(b-a)}{n})}{2}(f(x_{i-1})+f(x_i)\right) +\frac{\left(\frac{(b-a)}{n}\right)^2}{12}(f'(x_{i-1})-f'(x_i)) \\[10pt] = {} & \frac{(b-a)}{2n} \left( \sum_{i=1}^n((f(x_{i-1})+f(x_i)) +\frac{b-a}{6n}(f'(x_{i-1})-f'(x_i)) \right) \\[10pt] = {} & \frac{(b-a)}{2n} \left( \sum_{i=1}^n(f(x_{i-1})+f(x_i)) + \left(\frac{b-a}{6n}(f'(a)-f'(b))\right)\right). \end{align}

Here I'm stuck. I tried to simplify $f(x_{i-1})+f(x_i)$ but with no success.

Answer 1

$\begin{array}\\ \sum_{i=1}^n(f(x_{i-1})+f(x_i)) &=\sum_{i=1}^nf(x_{i-1})+\sum_{i=1}^nf(x_i)\\ &=\sum_{i=0}^{n-1}f(x_{i})+\sum_{i=1}^nf(x_i)\\ &=f(x_0)+\sum_{i=1}^{n-1}f(x_{i})+\sum_{i=1}^{n-1}f(x_i)+f(x_n)\\ &=f(x_0)+f(x_n)+2\sum_{i=1}^{n-1}f(x_{i})\\ \end{array} $