Take 2: When/Why are these equal? This didn't go right the first time, so I'm going to drastically rephrase the query.

As per this previous question, I am wondering if the two series $$\frac{f(a)+f(b)}{2}\frac{(b-a)}{1!}+\frac{f'(a)-f'(b)}{2}\frac{(b-a)^2}{2!}+\frac{f''(a)+f''(b)}{2}\frac{(b-a)^3}{3!}+\cdots$$ 

and $$\frac{f(a)+f(b)}{2}\frac{(b-a)}{1!}+\frac{f'(a)-f'(b)}{2^2}\frac{(b-a)^2}{2!}+\frac{f''(a)+f''(b)}{2^3}\frac{(b-a)^3}{3!}+\cdots$$ could possibly be equal. The only difference is the powers of $2$ in the denominator.

NOTE: the numerator is $f^{(k)}(a)+(-1)^kf^{(k)}(b)$ in general.
The identity you give is quite pretty! Yes they are equal. 

Let
$$\phi(t) = \sum_{n \ge 0}\frac{(b-a)^{n+1}}{(n+1)!}
\left( f^{(n)}(a) \cdot t^{n+1} + f^{(n)}(b) \cdot (-1)^n (1-t)^{n+1}\right).$$
The first series in question is $\frac12(\phi(1)+\phi(0))$ and the second is $\phi(\frac12)$. The "why" is that these are equal because, in fact, $\phi$ is constant:
$$\phi(t) = \int_a^bf(x) \, dx$$
for all $0 \le t \le 1$. The proof is by splitting the integral into two as
$$\int_a^bf(x) \, dx = \int_a^{a(1-t)+bt}f(x) \, dx + \int_{a(1-t)+bt}^bf(x) \, dx$$
and evaluating the first by expanding $f$ as a power series about $a$ and the second by expanding $f$ as a power series about $b$. You will get $\phi(t)$ as the result.

Specifically, the "when" is the condition that the power series expansions
$$f(x) = f(a)+f'(a)(x-a)+\frac{1}{2!}f''(a)(x-a)^2+\cdots$$
and
$$f(x) = f(b)+f'(b)(x-b)+\frac{1}{2!}f''(b)(x-b)^2+\cdots$$
are valid for all $x \in [a,b]$.

At the risk of sounding totally banal, I've basically just briefly rehashed (and slightly generalized) the derivation you linked to $-$ that post in itself already gives the proof.
Denote a primitive of $f$ by $F$ and add ${1\over2}\bigl(F(a)-F(b)\bigr)$ to both series. Then the first series becomes
$$\eqalign{{1\over2}\sum_{k=0}^\infty \bigl(F^{(k)}(a)-(-1)^kF^{(k)}(b)\bigr){(b-a)^k\over k!}&amp;={1\over2}\sum_{k=0}^\infty F^{(k)}(a){(b-a)^k\over k!} 
-{1\over2}\sum_{k=0}^\infty F^{(k)}(b){(a-b)^k\over k!}\cr
&amp;={1\over2}\bigl(F(b)-F(a)\bigr)\ ,\cr}$$
and similarly the second series becomes
$$-{1\over2}\bigl(F(a)-F(b)\bigr)+\sum_{k=0}^\infty \bigl(F^{(k)}(a)-(-1)^k F^{(k)}(b)\bigr)\&gt;{\bigl({b-a\over2}\bigr)^k\over k!}$$
$$={1\over2}\bigl(F(b)-F(a)\bigr)+F\left({a+b\over2}\right)-F\left({a+b\over2}\right)={1\over2}\bigl(F(b)-F(a)\bigr)\ .$$