Animated Solution for Mathematics - Binomial Theorem: In the binomial expansion of $(a-b{)}^n,n\ge 5$, the sum of the $5^{th}$ and $6^{th}$ terms is zero. Then $a/b$ equals

Unlocking the Binomial Mystery

Welcome, fellow traveler of the JEE landscape. Today, we are not just solving a problem; we are peeling back the layers of the Binomial Theorem.

When you see an expression like $(a-b{)}^n$, do not just see a formula. See a pattern, a rhythm, a dance of signs that unfolds as you expand it.

The problem asks us to find the ratio $\frac{a}{b}$ given that the sum of the $5^{th}$ and $6^{th}$ terms is zero. This is a classic JEE challenge—it tests your precision, your ability to handle signs, and your mastery of combination properties.

The General Term as Our Telescope

To look at specific terms in a binomial expansion, we need our most reliable tool: the general term formula. For any expansion of $(x+y{)}^n$, the $(r+1{)}^{th}$ term is given by:

In our case, $x=a$ and $y=-b$. This is where the magic—and the danger—begins.

We must carry that negative sign with us. It is not just a symbol; it is a mathematical instruction that dictates the sign of every term in our expansion.

The Sign Trap

Let us calculate the $5^{th}$ and $6^{th}$ terms. For the $5^{th}$ term ($T_5$), we set $r+1=5$, which means $r=4$.

Plugging this in, we get:

Since $4$ is an even power, the negative sign vanishes. Now, for the $6^{th}$ term ($T_6$), we set $r+1=6$, so $r=5$.

This gives us:

Because $5$ is an odd power, the negative sign persists. Do you see the beauty here? The $5^{th}$ term is positive, and the $6^{th}$ term is negative, reflecting the alternating nature of $(a-b{)}^n$.

The Algebraic Bridge

We are told that $T_5+T_6=0$. This is our bridge, implying that $T_5=-T_6$.

Substituting our expressions, we get:

The two negatives cancel out, leaving us with:

Now, we need to isolate $\frac{a}{b}$. We divide both sides by $a^{n-5}{b}^4$ to simplify the expression to:

The Elegant Shortcut

We are almost there. Rearranging for the ratio, we get:

As an elite JEE aspirant, you know the shortcut: the combination ratio property $\frac{{}^n{C}_r}{{}^n{C}_{r-1}}=\frac{n-r+1}{r}$. Here, $r=5$.

Substituting this, we get:

The JEE Mindset

And there it is. The final ratio is $\frac{n-4}{5}$.

We didn't just calculate; we navigated the structure of the binomial expansion. We respected the negative sign, we used the general term as a surgical tool, and we employed the combination ratio property to bypass tedious arithmetic.

This is the mindset that wins in JEE Advanced. Keep practicing, keep visualizing, and keep falling in love with the elegance of the math.