Animated Solution for Mathematics - Binomial Theorem: The coefficient of the middle term in the binomial expansion in powers of $x$ of $(1+\alpha x{)}^4$ and of $(1-\alpha x{)}^6$ is the same if $\alpha$ equals

Imagine you are standing before the vast, structured beauty of the Binomial Theorem. It is not merely a collection of formulas; it is the architecture of patterns.

Today, we are going to peel back the layers of two specific binomial expressions: $(1+\alpha x{)}^4$ and $(1-\alpha x{)}^6$. Our mission is to find the value of $\alpha$ that forces their middle terms to mirror each other perfectly.

This is a classic test of precision, where the smallest oversight—a missing negative sign—can lead us astray.

Locating the Center

First, we must orient ourselves. When we expand a binomial $(a+b{)}^n$, the number of terms is always $n+1$.

If $n$ is even, $n+1$ is odd, which gives us a single, unique middle term. For $(1+\alpha x{)}^4$, where $n=4$, the number of terms is 5. The middle term is the third one, $T_3$.

For $(1-\alpha x{)}^6$, where $n=6$, the number of terms is 7. The middle term is the fourth one, $T_4$. The formula for the position of the middle term is always $\frac{n}{2}+1$. It is the anchor of our calculation.

The Trap of the Negative Sign

Now, we apply the general term formula:

For our first expression, $(1+\alpha x{)}^4$, we seek $T_3$, so $r=2$. The coefficient is ${}^4{C}_2(\alpha {)}^2$. Since ${}^4{C}_2=6$, we get $6{\alpha }^2$.

But look at the second expression: $(1-\alpha x{)}^6$. This is where the trap lies. We seek $T_4$, so $r=3$.

The coefficient is ${}^6{C}_3(-\alpha {)}^3$. Notice the negative sign inside the parenthesis! Because the power $r=3$ is odd, the negative sign survives.

Thus, we get $-20{\alpha }^3$. If you had forgotten that negative sign, you would have been chasing a ghost.

The Algebraic Resolution

We are told these coefficients are equal. So, we set them in balance:

Now, we perform the final dance of algebra. Assuming $\alpha eq 0$, we divide both sides by $2{\alpha }^2$.

On the left, we are left with $3$. On the right, we are left with $-10\alpha$.

Thus, $3=-10\alpha$, which leads us to the elegant result:

We have navigated the geometry of the expansion and avoided the pitfalls of the negative sign. The beauty of this problem lies in its simplicity—once you respect the rules of the binomial expansion, the answer reveals itself.

Keep this rigor in your toolkit, and you will find that even the most intimidating problems are just puzzles waiting to be solved.