Animated Solution for Mathematics - Binomial Theorem: The sum of the series ${}^{20}{C}_0{-}^{20}{C}_1{+}^{20}{C}_2{-}^{20}{C}_3+....{+}^{20}{C}_{10}$ is

Welcome, student. Today, we are not just solving a math problem; we are uncovering a hidden symmetry in the world of binomial coefficients. When you look at a series like ${}^{20}{C}_0{-}^{20}{C}_1{+}^{20}{C}_2-\cdots {+}^{20}{C}_{10}$, it is easy to feel overwhelmed.

It looks like a chaotic mess of additions and subtractions. But I want you to pause and breathe. In mathematics, whenever you see a long, alternating sum of binomial coefficients, you are looking at a fragment of a much larger, more elegant structure.

Imagine you are standing on a vast plain, and you only see a small path in front of you. That is our series. To understand where we are, we need to see the whole landscape.

The master key to any binomial series is the Binomial Theorem. Recall the expansion of $(1-x{)}^n$:

Now, look at our problem. We have ${}^{20}{C}_0{-}^{20}{C}_1{+}^{20}{C}_2-\dots$. This is exactly what happens if we set $x=1$ and $n=20$.

When we do that, the left side becomes $(1-1{)}^{20}$, which is $0$. On the right side, we get the entire alternating sum from ${}^{20}{C}_0$ all the way to ${}^{20}{C}_{20}$. This tells us something profound: the sum of all these terms, from start to finish, is exactly $0$.

Now, let us look at the symmetry. One of the most beautiful properties in combinatorics is ${}^n{C}_r{=}^n{C}_{n-r}$. This means the coefficients are symmetric.

The first term, ${}^{20}{C}_0$, is equal to the last term, ${}^{20}{C}_{20}$. The second term, ${}^{20}{C}_1$, is equal to the second-to-last, ${}^{20}{C}_{19}$.

But what about the signs? We have an alternating series. The sign of the $r$-th term is determined by $(-1{)}^r$.

Because our power $n=20$ is an even number, the sign of the symmetric term $(-1{)}^{20-r}$ is identical to $(-1{)}^r$. This is the magic moment! The series is not just symmetric in value; it is symmetric in sign.

Let us define our series. We have the full sum equal to zero:

Let $S_{left}$ be the sum of the first ten terms, from ${}^{20}{C}_0$ to ${}^{20}{C}_9$. Let $S_{right}$ be the sum of the last ten terms, from ${}^{20}{C}_{11}$ to ${}^{20}{C}_{20}$. Because of the symmetry we just discussed, $S_{left}=S_{right}$.

Now, look at the equation again. We have the left part, the middle term ${}^{20}{C}_{10}$, and the right part. So:

Since $S_{left}=S_{right}$, we can write this as:

This is the heart of the problem. We have compressed this massive, intimidating series into a simple linear equation. We can easily solve for $S_{left}$:

Don't Fall into the Trap

We are almost there, but do not celebrate just yet! The question asks for the sum up to ${}^{20}{C}_{10}$. Our $S_{left}$ only goes up to ${}^{20}{C}_9$.

We must add the middle term, ${}^{20}{C}_{10}$, back into our sum to get the final answer:

Substituting our value for $S_{left}$:

And what is one minus a half? It is a half. So, the final result is:

See how elegant that was? We didn't need to calculate massive factorials. We didn't need to grind through twenty terms. We used the symmetry of the binomial coefficients and the power of the binomial expansion to let the math do the work for us.