I'm trying to derive the Euler Maclaurin formula following steps provided in Bender and Orszag's Advanced Mathematical Methods for Scientists and Engineers problem 6.88 page 315.
Given a sum of the form:
(1)$$F(n) = \sum_{k=0}^{n}f(k)$$
I've shown that
(2)$$F(n) = \frac{1}{2}[f(0)+f(n)] + \int_{0}^{n}f(t)dt + \int_{0}^{n}B_{1}(t+[t])f'(t)dt $$
where $B_{1}(t)$ is the first Bernoulli polynomial and $[t]$ is the floor function.
I've also shown that
(3)$$B_{n}'(x) = nB_{n-1}(x)$$ and $$B_{n}(0) = B_{n}(1)$$ for all $n\ge2$
Now I'm trying to show through integration by parts of eq. (2) m times, we will obtain
$$F(n) = \frac{1}{2}[f(0)+f(n)] + \int_{0}^{n}f(t)dt +\sum_{j=1}^{m}(-1)^{j+1}\frac{B_{j+1}}{(j+1)!}[f^{(j)}(n)-f^{(j)}(0)] + \frac{(-1)^m}{(m+1)!}\int_{0}^{n}B_{m+1}(t+[t])f'(t)dt $$
I'm trying to derive the above statement through induction on $m$ and applying integration by parts (IBP) on $\int_{0}^{n}B_{1}(t+[t])f'(t)dt$ where $m$ is the number of times we apply IBP We apply integration by parts:
$$\int_{0}^{n}B_{1}(t+[t])f'(t)dt = \int_{0}^{n}\frac{1}{2}B_{2}'(t+[t])f'(t)dt$$ $$= \sum_{k=0}^{n-1}\int_{k}^{k+1}\frac{1}{2}B_{2}'(t+k)f'(t)dt$$ $$ = \sum_{k=0}^{n-1}\frac{1}{2}B_{2}(t+k)f'(t)]^{k+1}_{k} - \int_{0}^{n}B_{2}(t+[t])f''(t)dt $$ $$ = \sum_{k=0}^{n-1}\frac{1}{2}[B_{2}(2k+1)f'(k+1)-B_{2}(2k)f'(k)] - \int_{0}^{n}B_{2}(t+[t])f''(t)dt $$
This is were I'm stuck, I'm unsure how to obtain this identity:
$$\sum_{k=0}^{n-1}\frac{1}{2}[B_{2}(2k+1)f'(k+1)-B_{2}(2k)f'(k)] = \frac{1}{2} B_{2}[f'(n)-f'(0)]$$
any hints would be greatly appreciated.