Well, that is not quite true...
$\displaystyle \sum\limits_{k = 1}^{100} {k\left[ {\frac{{f\left( {a + \tfrac{k}{n}} \right) - f(a)}}{{\tfrac{k}{n}}}} \right]} = n\left[ {\sum\limits_{k = 1}^{100} {\left( {f\left( {a + \frac{k}{n}} \right) - f(a)} \right)} } \right] = n\left[ {\left( \sum\limits_{k = 1}^{100} {f\left( {a + \frac{k}{n}} \right) \right) - 100f(a)} } \right] $
So, your RHS is correct, but the LHS is not (misplaced parenthesis).
I do not quite understand how you went from $\displaystyle n\left[ {\left( \sum\limits_{k = 1}^{100} {f\left( {a + \frac{k}{n}} \right) \right) - 100f(a)} } \right] $ to $\displaystyle n\left[ {\sum\limits_{k = 1}^{100} {\left( {f\left( {a + \frac{k}{n}} \right) - f(a)} \right)} } \right]$ and incorporated the $\displaystyle 100f(a)$ term into the summation. Also how does the outlying $\displaystyle n$ disappear?
Assuming that all of this can be reduced down to $\displaystyle \sum\limits_{k = 1}^{100} {k\left[ {\frac{{f\left( {a + \tfrac{k}{n}} \right) - f(a)}}{{\tfrac{k}{n}}}} \right]}$ would I now consider?
$\displaystyle \lim_{n\to\infty}\sum\limits_{k = 1}^{100} {k\left[ {\frac{{f\left( {a + \tfrac{k}{n}} \right) - f(a)}}{{\tfrac{k}{n}}}} \right]}$
So here if you look at individual parts of the series you have: $\displaystyle f\left( {a + \tfrac{k}{n}} \right) \equiv f(a)$ since $\displaystyle \frac{k}{n}$ would approach 0 as n approaches infinity. It would follow similarly for the denominator and would cause the whole series to converge to 0?
How does the fact that f(x) is differentiable at a point x=a and f'(a) = 1 come into play?
Let's write out the summation $\displaystyle n\left[ {\sum\limits_{k = 1}^{100} {\left( {f\left( {a + \frac{k}{n}} \right) - f(a)} \right)} } \right]$:
$\displaystyle n \left[ \left( f\left(a + \dfrac{1}{n} \right) - f(a) \right) + \left( f\left(a + \dfrac{2}{n} \right) - f(a) \right) + \ldots + \left( f\left(a + \dfrac{100}{n} \right) - f(a) \right) \right]$
So, how many times are you subtracting $\displaystyle f(a)$? Since there are 100 terms, you are subtracting it 100 times.
The outlying $\displaystyle n$ does not disappear. When you divide by a fraction, that is the same as multiplying by its reciprocal. So, inside the summation we have:
$\displaystyle \begin{align*}k\left[ \dfrac{f\left( a + \tfrac{k}{n} \right) - f(a)}{\tfrac{k}{n}} \right] & = k \dfrac{n}{k}\left[ f\left( a + \dfrac{k}{n} \right) - f(a) \right] \\ & = n\left[ f\left( a + \dfrac{k}{n} \right) - f(a) \right]\end{align*}$
Now, you can factor $\displaystyle n$ from every term, bringing it outside the summation (you can do this since $\displaystyle n$ does not depend on $\displaystyle k$).
Use your limit laws. The limit of a sum is the sum of the limits, so long as those limits exist (and so long as the sum is finite... infinite sums require more work). So:
$\displaystyle \lim_{n\to\infty}\sum\limits_{k = 1}^{100} {k\left[ {\frac{{f\left( {a + \tfrac{k}{n}} \right) - f(a)}}{{\tfrac{k}{n}}}} \right]} = \sum\limits_{k = 1}^{100} {k\left[ \lim_{n\to\infty} {\frac{{f\left( {a + \tfrac{k}{n}} \right) - f(a)}}{{\tfrac{k}{n}}}} \right]}$
As $\displaystyle n\to \infty$, $\displaystyle \dfrac{k}{n} \to 0$. So, we can rewrite this limit:
$\displaystyle \sum\limits_{k = 1}^{100} {k\left[ \lim_{n\to\infty} {\frac{{f\left( {a + \tfrac{k}{n}} \right) - f(a)}}{{\tfrac{k}{n}}}} \right]} = \sum\limits_{k = 1}^{100} {k\left[ \lim_{\tfrac{k}{n}\to 0} {\frac{{f\left( {a + \tfrac{k}{n}} \right) - f(a)}}{{\tfrac{k}{n}}}} \right]}$
Next, you asked about how you use $\displaystyle f'(a) = 1$. Use the definition of the derivative. For a general function $\displaystyle f(x)$, you know $\displaystyle f'(a) = ???$ (Hint: When you write it out, replace $\displaystyle h$ by $\displaystyle \tfrac{k}{n}$)
For your question about the series converging to 0, no. You have the numerator and denominator approaching zero. 0/0 is not defined, so you need to simplify the expression before you can take the limit.
Ok I think I have this entirely put together, can you tell me if this looks right?
$\displaystyle \lim_{n\to\infty}n[f(a + \frac{1}{n}) + f(a + \frac{2}{n}) + ... + f(a + \frac{100}{n}) - 100f(a)\right]$
$\displaystyle = \lim_{n\to\infty}n\left[(\sum\limits_{k=1}^{100}f(a + \frac{k}{n})) - 100f(a)\right]$
$\displaystyle = \lim_{n\to\infty}n\left[\sum\limits_{k=1}^{100}(f(a + \frac{k}{n}) - f(a))\right]$
$\displaystyle = \lim_{n\to\infty}\sum\limits_{k=1}^{100}k\frac{n}{ k}\left[f(a + \frac{k}{n}) - f(a))\right]$
$\displaystyle = \lim_{n\to\infty}\sum\limits_{k=1}^{100}k\left[\frac{f(a + \frac{k}{n}) - f(a)}{\frac{k}{n}}\right]$
Now, since the limit of a sum is the sum of the limits, we can write:
$\displaystyle \lim_{n\to\infty}\sum\limits_{k = 1}^{100} {k\left[ {\frac{{f\left( {a + \tfrac{k}{n}} \right) - f(a)}}{{\tfrac{k}{n}}}} \right]} = \sum\limits_{k = 1}^{100} {k\left[ \lim_{n\to\infty} {\frac{{f\left( {a + \tfrac{k}{n}} \right) - f(a)}}{{\tfrac{k}{n}}}} \right]}$
As $\displaystyle n\to \infty$, $\displaystyle \dfrac{k}{n} \to 0$. So, we can rewrite this limit:
$\displaystyle \sum\limits_{k = 1}^{100} {k\left[ \lim_{n\to\infty} {\frac{{f\left( {a + \tfrac{k}{n}} \right) - f(a)}}{{\tfrac{k}{n}}}} \right]} = \sum\limits_{k = 1}^{100} {k\left[ \lim_{\tfrac{k}{n}\to 0} {\frac{{f\left( {a + \tfrac{k}{n}} \right) - f(a)}}{{\tfrac{k}{n}}}} \right]}$
Note: the definition of a derivative is $\displaystyle f'(a) = \lim_{h\to\0} \frac{f(a+h) - f(a)}{h}.$
We have this here in our modified limit: $\displaystyle \sum\limits_{k = 1}^{100} {k\left[ \lim_{\tfrac{k}{n}\to 0} {\frac{{f\left( {a + \tfrac{k}{n}} \right) - f(a)}}{{\tfrac{k}{n}}}} \right]}$.
So we can write $\displaystyle \sum\limits_{k = 1}^{100} {k\left[ \lim_{\tfrac{k}{n}\to 0} {\frac{{f\left( {a + \tfrac{k}{n}} \right) - f(a)}}{{\tfrac{k}{n}}}} \right]} = \sum\limits_{k = 1}^{100}kf'(a) = \sum\limits_{k = 1}^{100}k = 5050. $
Is that straightforward and correct?