Quick ways for approximating $\sum_{k=a_1}^{k=a_2}C_{100}^k(\frac{1}{2})^k(\frac{1}{2})^{100-k}$? Consider the following problem:

A fair coin is to be tossed 100 times, with each toss resulting in a head or a tail. Let
  $$H:=\textrm{the total number of heads}$$
  and 
  $$T:=\textrm{the total number of tails},$$
  which of the following events has the greatest probability?
A. $H=50$
B. $T\geq 60$
C. $51\leq H\leq 55$
D. $H\geq 48$ and $T\geq 48$
E. $H\leq 5$ or $H\geq 95$

What I can think is the direct calculation:
$$P(a_1\leq H\leq a_2)=\sum_{k=a_1}^{k=a_2}C_{100}^k(\frac{1}{2})^k(\frac{1}{2})^{100-k}$$
Here is my question:
Is there any quick way to solve this problem except the direct calculation?

Here is a very elementary way of estimating these probabilities.  Observe that the distribution of $H$ is very similar to a normal distribution with mean $50$ and standard deviation $\sigma = 5$.  In particular, we should have
$$
P (|H-50| \leq \sigma) \;\approx\; 68\% \qquad\text{and}\qquad P(|H-50| \leq 2\sigma) \;\approx\; 95\%
$$
As mixedmath pointed out, the only viable answers are B, D, and E.  We can estimate the probabilities of these events as follows:
B. $P(H \geq 60) \;=\; P(H \geq 50 + 2\sigma)$, which should be on the order of 2.5%.
D. $P(48\leq H \leq 52) \;=\; P(|H-50| \leq \sigma/2)$, so this should be something like 40%.
E. $P(H\leq 5\text{ or }H \geq 95) \;=\; P(|H-50| \geq 9\sigma)$, so this should be really small.
Thus (D) is the correct answer.
