If the coefficient of x¹⁵ in the expansion of (ax³ + 1/bx^(1/3))¹⁵ is equal to the coefficient of x⁻

Q: Wait, why does the negative sign in the second expansion matter? Can't I just ignore it?

Ah, the classic trap! In the expansion $(ax^{1/3} - \frac{1}{bx^3})^{15}$, that negative sign is part of the second term. It becomes $(-1)^k$. If $k$ is odd, the term is negative; if $k$ is even, it's positive. Ignoring it is a recipe for disaster because it changes the sign of your final coefficient!

Q: What if I get a non-integer value for $r$?

If you get a non-integer for $r$, it means the term you are looking for simply does not exist in the expansion. But in JEE problems, we usually design them so the term exists. If you get a fraction, re-check your exponent arithmetic—you likely missed a term or made a sign error in your power summation.

Q: Is there a shortcut for $\binom{15}{9}$ and $\binom{15}{6}$?

Absolutely! The property $\binom{n}{r} = \binom{n}{n-r}$ is your best friend. Since $15-9 = 6$, $\binom{15}{9}$ is identical to $\binom{15}{6}$. Never calculate these large factorials unless absolutely necessary; look for symmetry first!

Question

If the coefficient of $x^{15}$ in the expansion of $(ax^3 + \frac{1}{bx^{1/3}})^{15}$ is equal to the coefficient of $x^{-15}$ in the expansion of $(ax^{1/3} - \frac{1}{bx^3})^{15}$, where $a$ and $b$ are positive real numbers, then for each such ordered pair $(a, b)$ :

Options:

(A) $a = b$

(B) $ab = 1$

(C) $a = 3b$

(D) $ab = 3$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, aspiring engineers! Today, we are embarking on a journey through the elegant world of the Binomial Theorem. Our mission is to find the hidden relationship between two positive real numbers, $a$ and $b$, by equating the coefficients of specific powers of $x$ in two distinct expansions.

### Phase 1: The First Expansion

We begin with the first expression: $(ax^3 + \frac{1}{bx^{1/3}})^{15}$. To find the coefficient of $x^{15}$, we invoke the general term formula:

$$T_{r+1} = \binom{15}{r} (ax^3)^{15-r} \left(\frac{1}{bx^{1/3}}\right)^r$$

Here, $r$ is our index, ranging from $0$ to $15$. We separate the constants from the variables to obtain:

$$\binom{15}{r} a^{15-r} b^{-r} \cdot x^{3(15-r)} \cdot x^{-r/3}$$

By applying the laws of exponents, the power of $x$ becomes $45 - 3r - \frac{r}{3}$. Setting this exponent equal to $15$, we solve the equation:

$$45 - \frac{10r}{3} = 15 \Rightarrow 30 = \frac{10r}{3} \Rightarrow r = 9$$

The resulting coefficient for the first expression is $\binom{15}{9} a^6 b^{-9}$.

### Phase 2: The Second Expansion

Now, we turn to the second expression: $(ax^{1/3} - \frac{1}{bx^3})^{15}$. We use a new index, $k$, to define the general term:

$$T_{k+1} = \binom{15}{k} (ax^{1/3})^{15-k} \left(-\frac{1}{bx^3}\right)^k$$

Note the negative sign, which introduces a factor of $(-1)^k$. Isolating the powers of $x$, we get $x^{(15-k)/3} \cdot x^{-3k}$, resulting in an exponent of $5 - \frac{k}{3} - 3k$.

Setting this exponent equal to $-15$, we solve for $k$:

$$5 - \frac{10k}{3} = -15 \Rightarrow 20 = \frac{10k}{3} \Rightarrow k = 6$$

Since $k=6$ is even, $(-1)^6 = 1$. The coefficient simplifies to $\binom{15}{6} a^9 b^{-6}$.

### Phase 3: The Symmetry of Success

We now equate the two coefficients:

$$\binom{15}{9} a^6 b^{-9} = \binom{15}{6} a^9 b^{-6}$$

Because $\binom{15}{9} = \binom{15}{6}$, these terms cancel out perfectly. We are left with the following relationship:

$$a^6 b^{-9} = a^9 b^{-6}$$

Rearranging the terms, we divide both sides by $a^6$ and $b^{-6}$:

$$\frac{a^6}{a^9} = \frac{b^{-6}}{b^{-9}} \Rightarrow a^{-3} = b^3$$

This implies $1 = a^3 b^3$, or $(ab)^3 = 1$. Since $a$ and $b$ are positive real numbers, we conclude that the final relationship is:

**$ab = 1$**

Select Answer:

Visualized Solution

Problem Overview

General Term of $(a x^{3} + \frac{1}{b x ^{\frac{1}{3}}})^{15}$

Simplifying the Exponent of $x$

Equating Exponent to $15$

Extracting the First Coefficient

General Term of $(a x^{\frac{1}{3}} - \frac{1}{b x ^{3}})^{15}$

Simplifying the Exponent of $x$

Equating Exponent to $- 15$

Extracting the Second Coefficient

Applying the Given Condition

Solving for $a$ and $b$

The Sigma Insight: General Term and Middle Term

Analyzing the Setup

Phase 1

Phase 2

Phase 3

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Visualized Solution

Problem Overview

General Term of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } (ax3+bx31​1​)15

Simplifying the Exponent of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x

Equating Exponent to .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 15

Extracting the First Coefficient

General Term of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } (ax31​−bx31​)15

Simplifying the Exponent of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x

Equating Exponent to .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } −15

Extracting the Second Coefficient

Applying the Given Condition

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } b

The Sigma Insight: General Term and Middle Term

Analyzing the Setup

Phase 1

Phase 2

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