Assert the range of a binomial coefficient divided by power of a number

Question

I found this question in a previous year post graduate entrance exam for mathematics.The question was
What is the range of \begin{equation*} \frac{200 \choose 100}{4^{100}} \end{equation*}

The choices were
\begin{align*} [\frac{3}{4}, 1) && \text{or} &&(0, \frac{1}{2}) && \text{or} && [1, \infty) && \text{or} && [\frac{1}{2}, \frac{3}{4}) \end{align*}

Answer 1

Hint:

Compare $\binom{200}{100}$ with $\sum_{k=0}^{200}\binom{200}k$.

This in the understanding that: $$\sum_{k=0}^{200}\binom{200}k=2^{200}=4^{100}$$

Answer 2

$$\binom{n}{0}+\binom{n}{1}+\binom{n}{2}+...+\binom{n}{n}=2^n$$ so you can rewrite $$\begin{equation*} \frac{200 \choose 100}{4^{100}}=\frac{200 \choose 100}{2^{200}}=\frac{200 \choose 100}{\binom{200}{0}+\binom{200}{1}+\binom{200}{2}+...+\binom{200}{100}+...+\binom{200}{199}+\binom{200}{200}}<<1 \end{equation*}$$ you can make a sence by looking at pascal's triangle $$1\\1\space 1\\1 \space 2 \space 1 \\ 1 \space 3 \space 3 \space 1\\1\space 4\space 6 \space 4 \space 1\\1\space 5 \space 10 \space 10 \space 5\space 1\\1\space 6 \space 15 \space 20 \space 15\space 6 \space 1$$ for example at $n=4\to \frac{6}{16}<\frac8{16}$ and $n=6\to \frac{20}{64}<\frac 13$ and so on

Answer 3

Write the expression as $$\frac {200!}{(2^{100}\cdot 100!)(2^{100}\cdot 100!)}$$ The first term in the denominator cancels all the even factors in the numerator, leaving us with $$\left(\frac{199}{200}\right)\left(\frac{197}{198}\right)\left(\frac{195}{196}\right)\ldots\left(\frac{1}{2}\right)$$ The last factor is $\frac 12$ and all the rest are less than $1$, so the product is less than $\frac 12$