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?
 A: Yes.  In the following, which uses (almost) nothing beyond what is already in the problem statement, the calculations involve only simple arithmetic with one-digit numbers (and $10$) and easy estimates involving fractions of two-digit numbers: the stuff of mental arithmetic.
Let $P(k)$ represent the probability of $k$ heads.  From the (intuitively obvious) facts that (i) $P(k) \gt 0$ for $0 \le k \le 100$, (ii) $P(k)$ increases from $k=0$ to $k=50$ and then decreases from $k=50$ to $k=100$, and (iii) $P(k) = P(100-k)$, we easily establish the inequalities
$$D \gt C, D \gt A, B \gt E.$$
I claim that actually $A \gt B$ (i.e., the chance of exactly 50 heads exceeds the chance of 60 or more tails), which establishes $D$ as the answer.  To see this, look at the relative probabilities.  They all have a common factor of $100!/2^{100}$ which we can ignore, focusing on the binomial coefficients that are left.  Now a series of simple estimates establishes
$$P(40) / P(50) = \frac{50}{60} \frac{49}{59} \cdots \frac{41}{51} \lt \left(\frac{5}{6}\right)^{10} \lt \frac{1}{1 + 10(1/5)} = \frac{1}{3}.$$
(The ratio actually is less than $1/7$.)  Moreover, 
$$P(39) / P(50) = \frac{40}{61} P(40)/P(50) \lt \frac{2}{3} P(40)/P(50).$$
Continuing in like vein we see that the chance of $A$ relative to that of $B$, $\left(P(0) + P(1) + \cdots + P(40)\right)/P(50)$, is dominated by a geometric series with starting term $P(40)/P(50)$ and common ratio $2/3$.  Therefore its sum is less than $1/3 (1 - 2/3)^{-1} = 1.$  This proves the claim.
A: 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.
A: 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.
A: In short: no. But you can cut a few things out immediately. For example, A is nonsense. A is eaten by C. And C is eaten by D.
So you need only to check B, D, and E. Of course, depending on your intuition, you might have a 'feel' for how unlikely E is as well. But that isn't as certain.
