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?

Chebyshev's inequality, combined with mixedmath's and some other observations, shows that the answer has to be D without doing the direct calculations. 
First, rewrite D as $48 \leq H \leq 52$.  A is a subset of D, and because the binomial distribution with $n = 100$ and $p = 0.5$ is symmetric about $50$, C is less likely than D.  So, as mixedmath notes, A and C can be ruled out.
Now, estimate the probability of D.  We have $P(H = 48) = \binom{100}{48} 2^{-100} > 0.07$.  Since $H = 48$ and $H=52$ are equally probable and are the least likely outcomes in D, $P(D) > 5(0.07) = 0.35$. 
Finally, $\sigma_H = \sqrt{100(0.5)(0.5)} = 5$.  So the two-sided version of Chebyshev says that $P(E) \leq \frac{1}{9^2} = \frac{1}{81}$, since E asks for the probability that $H$ takes on a value 9 standard deviations away from the mean.  The one-sided version of Chebyshev says that $P(B) \leq \frac{1}{1+2^2} = \frac{1}{5}$, since B asks for the probability that $H$ takes on a value 2 standard deviations smaller than the mean. 
So D must be the most probable event.

Added: OP asks for more on why $P(C) < P(D)$.  Since the binomial($100,50$) distribution is symmetric about $50$, $P(H = i) > P(H = j)$ when $i$ is closer to $50$ than $j$ is.  Thus $$P(C) = P(H = 51) + P(H = 52) + P(H = 53) + P(H = 54) + P(H = 55)$$ $$< P(H = 50) + P(H=51) + P(H = 49) + P(H = 52) + P(H = 48) = P(D),$$ by directly comparing probabilities.
