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.

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
$$\eqalign{{1\over2}\sum_{k=0}^\infty \bigl(F^{(k)}(a)-(-1)^kF^{(k)}(b)\bigr){(b-a)^k\over k!}&={1\over2}\sum_{k=0}^\infty F^{(k)}(a){(b-a)^k\over k!} 
-{1\over2}\sum_{k=0}^\infty F^{(k)}(b){(a-b)^k\over k!}\cr
&={1\over2}\bigl(F(b)-F(a)\bigr)\ ,\cr}$$
and similarly the second series becomes
$$-{1\over2}\bigl(F(a)-F(b)\bigr)+\sum_{k=0}^\infty \bigl(F^{(k)}(a)-(-1)^k F^{(k)}(b)\bigr)\>{\bigl({b-a\over2}\bigr)^k\over k!}$$
$$={1\over2}\bigl(F(b)-F(a)\bigr)+F\left({a+b\over2}\right)-F\left({a+b\over2}\right)={1\over2}\bigl(F(b)-F(a)\bigr)\ .$$
