In the binomial expansion of (a - b)ⁿ, n ≥ 5, the sum of 5th and 6th terms is zero, then a/b equals

Q: Wait, why do we use $r=4$ for the $5^{th}$ term and $r=5$ for the $6^{th}$ term? Shouldn't it be $r=5$ and $r=6$?

That is a classic point of confusion! Remember that the general term formula is $T_{r+1} = \binom{n}{r} a^{n-r} b^r$. The index $r$ starts from $0$. So, for the first term ($T_1$), $r=0$. For the second term ($T_2$), $r=1$. Following this pattern, for the $k^{th}$ term, $r$ must be $k-1$. Thus, for the $5^{th}$ term, $r=4$, and for the $6^{th}$ term, $r=5$. Keep this index shift in mind, and you will never go wrong!

Q: Why does the negative sign disappear in the $5^{th}$ term but stay in the $6^{th}$ term?

It all comes down to the parity of the exponent! In our expansion, the term involving $b$ is $(-b)^r$. When $r=4$ (for the $5^{th}$ term), the exponent is even, so $(-b)^4 = b^4$. When $r=5$ (for the $6^{th}$ term), the exponent is odd, so $(-b)^5 = -b^5$. Always check if the power of the negative term is even or odd—it is a simple detail that changes the entire sign of your result.

Q: Is there a faster way to calculate $\frac{\binom{n}{5}}{\binom{n}{4}}$ without expanding all the factorials?

Absolutely! Expanding factorials like $\frac{n!}{5!(n-5)!} \div \frac{n!}{4!(n-4)!}$ is a recipe for algebraic errors. Instead, use the powerful ratio identity: $\frac{\binom{n}{r}}{\binom{n}{r-1}} = \frac{n-r+1}{r}$. By plugging in $r=5$, you get $\frac{n-5+1}{5} = \frac{n-4}{5}$ instantly. This is a standard JEE trick to save precious time during the exam.

Question

In the binomial expansion of $(a - b)^n, n \ge 5$, the sum of $5^{th}$ and $6^{th}$ terms is zero, then $a/b$ equals

Options:

(A) $\frac{n - 5}{6}$

(B) $\frac{n - 4}{5}$

(C) $\frac{5}{n - 4}$

(D) $\frac{6}{n - 5}$

Sigma Percentile · Accepted Answer

### The Elegance of Binomial Symmetry: Decoding $(a - b)^n$

Welcome, aspiring engineers! Today, we are going to peel back the layers of a classic Binomial Theorem problem. It is not just about crunching numbers; it is about understanding the hidden symmetry in algebraic expansions.

We are given the expansion of $(a - b)^n$ where $n \ge 5$, and we are told that the sum of the $5^{th}$ and $6^{th}$ terms is zero. Our goal is to find the ratio $\frac{a}{b}$.

### Phase 1: The General Term Strategy

The heartbeat of any binomial problem is the general term formula. For any expansion $(x + y)^n$, the general term is:

$$T_{r+1} = \binom{n}{r} x^{n-r} y^r$$

In our specific case, we are dealing with $(a - b)^n$. This means our $x$ is $a$ and our $y$ is $-b$. It is crucial to treat that negative sign as part of the $y$ term.

So, our general term becomes:

$$T_{r+1} = \binom{n}{r} a^{n-r} (-b)^r$$

This is our primary weapon.

### Phase 2: The Trap of the Negative Sign

Now, let us find our specific terms. For the $5^{th}$ term, $T_5$, we set $r = 4$. Substituting this into our formula, we get:

$$T_5 = \binom{n}{4} a^{n-4} (-b)^4 = \binom{n}{4} a^{n-4} b^4$$

Since $4$ is an even number, the negative sign vanishes.

Now, for the $6^{th}$ term, $T_6$, we set $r = 5$. This gives us:

$$T_6 = \binom{n}{5} a^{n-5} (-b)^5 = -\binom{n}{5} a^{n-5} b^5$$

Because $5$ is an odd number, the negative sign persists. This is where many students stumble—forgetting that the sign of the term depends entirely on the parity of $r$.

### Phase 3: The Algebraic Dance

The problem gives us a beautiful condition: $T_5 + T_6 = 0$. Substituting our expressions, we have:

$$\binom{n}{4} a^{n-4} b^4 - \binom{n}{5} a^{n-5} b^5 = 0$$

To solve for $\frac{a}{b}$, we shift the negative term to the right:

$$\binom{n}{4} a^{n-4} b^4 = \binom{n}{5} a^{n-5} b^5$$

Now, we group the variables. Dividing both sides by $a^{n-5}$ and $b^4$, we get:

$$\frac{a^{n-4}}{a^{n-5}} = \frac{\binom{n}{5}}{\binom{n}{4}} \cdot \frac{b^5}{b^4}$$

Using the laws of exponents, $a^{(n-4)-(n-5)} = a^1 = a$ and $\frac{b^5}{b^4} = b^{5-4} = b$. Thus, we arrive at:

$$a = \frac{\binom{n}{5}}{\binom{n}{4}} b \Rightarrow \frac{a}{b} = \frac{\binom{n}{5}}{\binom{n}{4}}$$

### Phase 4: The Shortcut to Victory

We could expand the binomial coefficients using factorials, but that is the long road. Instead, we use the ratio identity:

$$\frac{\binom{n}{r}}{\binom{n}{r-1}} = \frac{n-r+1}{r}$$

Here, $r=5$. Substituting this into our identity, we get:

$$\frac{\binom{n}{5}}{\binom{n}{4}} = \frac{n-5+1}{5} = \frac{n-4}{5}$$

And there it is! The ratio is:

$$\mathbf{\frac{a}{b} = \frac{n-4}{5}}$$

This problem is a perfect example of how identifying the general term and using properties of binomial coefficients can turn a daunting algebraic expression into a simple, elegant result. Keep practicing, stay curious, and remember: in mathematics, the most complex problems often have the most beautiful, simple solutions.

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

Select Answer:

Visualized Solution

Problem Statement

General Term Formula

Finding the $5^{t h}$ Term ( $T_{5}$ )

Finding the $6^{t h}$ Term ( $T_{6}$ )

Applying the Given Condition

Rearranging the Equation

Grouping $a$ and $b$ Terms

Simplifying Powers of $a$ and $b$

The Binomial Coefficient Ratio Formula

Calculating the Ratio

Final Result

The Sigma Insight: General Term and Middle Term

The Elegance of Binomial Symmetry

Phase 1

Phase 2

Phase 3

Phase 4

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Animated Solution for Mathematics - Binomial Theorem: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } In the binomial expansion of (a−b)n,n≥5, the sum of 5th and 6th terms is zero, then a/b equals