Animated Solution for Mathematics - Binomial Theorem: The ratio of the coefficient of the middle term in the expansion of $(1+x{)}^{20}$ and the sum of the coefficients of two middle terms in expansion of $(1+x{)}^{19}$ is

A Binomial Journey

Imagine standing before the vast, shimmering expanse of Pascal's Triangle. It is not merely a collection of numbers; it is the map of all possible choices, the very architecture of combinations.

Today, we are going to explore a problem that might look like a tedious exercise in arithmetic, but is actually a beautiful demonstration of symmetry in the Binomial Theorem. We are comparing the middle terms of $(1+x{)}^{20}$ and $(1+x{)}^{19}$.

The Even Power Case ($n=20$)

First, let us focus on the expansion of $(1+x{)}^{20}$. Here, the power $n$ is $20$, which is an even number.

The Binomial Theorem tells us that the expansion of $(1+x{)}^n$ contains $n+1$ terms. So, for $n=20$, we have $21$ terms.

Because $21$ is an odd number, there is a single, unique term sitting right in the center. To find its position, we use the formula $\frac{n}{2}+1$.

Substituting $n=20$, we get $\frac{20}{2}+1=11$. Thus, the $11^{\text{th}}$ term is our middle term.

Using the general term formula $T_{r+1}{=}^n{C}_r{x}^r$, we see that for the $11^{\text{th}}$ term, $r=10$. Therefore, the coefficient of this middle term is ${}^{20}{C}_{10}$.

The Odd Power Case ($n=19$)

Now, shift your gaze to the second expansion: $(1+x{)}^{19}$. Here, the power $n$ is $19$, an odd number.

The total number of terms is $19+1=20$. Since $20$ is an even number, there is no single middle term; instead, we have two middle terms that share the center stage.

The positions are given by $\frac{n+1}{2}$ and $\frac{n+1}{2}+1$. For $n=19$, this gives us $\frac{20}{2}=10$ and $11$.

So, the $10^{\text{th}}$ and $11^{\text{th}}$ terms are our middle terms. The coefficients are ${}^{19}{C}_9$ (for the $10^{\text{th}}$ term) and ${}^{19}{C}_{10}$ (for the $11^{\text{th}}$ term).

The problem asks for the sum of these two coefficients: ${}^{19}{C}_9{+}^{19}{C}_{10}$.

The Bridge of Pascal's Identity

This is where the magic happens. You might be tempted to calculate these values, but pause.

Look at the expression ${}^{19}{C}_9{+}^{19}{C}_{10}$. It is the classic form of Pascal's Identity:

By applying this identity with $n=19$ and $r=10$, the sum ${}^{19}{C}_9{+}^{19}{C}_{10}$ collapses perfectly into ${}^{20}{C}_{10}$.

We have arrived at the climax of our journey. We need the ratio of the coefficient from the first expansion to the sum of the coefficients from the second.

We found the first to be ${}^{20}{C}_{10}$, and we just proved the sum of the second is also ${}^{20}{C}_{10}$. When we divide them, we get:

It is not a coincidence; it is the inherent symmetry of the binomial coefficients. The final answer is 1.