If the co-efficient of x⁹ in (αx³ + 1/βx)¹¹ and the co-efficient of x⁻⁹ in (αx - 1/βx³)¹¹ are equal,

Q: Wait, why is the second term negative in the second expansion? Does it change the coefficient?

That is a fantastic observation! In the second expansion, $(\alpha x - \frac{1}{\beta x^3})^{11}$, the second term is indeed negative. When we raise this to the power of $k$ in the general term formula, we get $(-\frac{1}{\beta x^3})^k$, which is $(-1)^k \cdot (\frac{1}{\beta x^3})^k$. Since $k=5$ in our solution, $(-1)^5$ is $-1$. This negative sign is crucial—if you miss it, your final answer will be wrong. Always keep track of the sign of the terms inside the binomial!

Q: How do I know if I should use $r$ or $k$ for the indices?

You are absolutely right to be cautious! Using different variables like $r$ and $k$ is a best practice in mathematics to avoid confusion. Since we are dealing with two separate expansions, the index for the term in the first expansion has no relation to the index in the second. Using distinct variables ensures you don't accidentally equate them or mix up their values.

Q: Can I just calculate $\binom{11}{6}$ and $\binom{11}{5}$? They seem like big numbers.

You could, but it is unnecessary and prone to calculation errors. The beauty of the Binomial Theorem lies in its properties. Recognizing that $\binom{n}{r} = \binom{n}{n-r}$ is a hallmark of a seasoned problem solver. It saves time and reduces the complexity of the algebra. Always look for these symmetries before diving into heavy arithmetic!

Question

If the co-efficient of $x^9$ in $(\alpha x^3 + \frac{1}{\beta x})^{11}$ and the co-efficient of $x^{-9}$ in $(\alpha x - \frac{1}{\beta x^3})^{11}$ are equal, then $(\alpha\beta)^2$ is equal to ______.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

We are tasked with finding the value of $(\alpha\beta)^2$ given two binomial expansions: $(\alpha x^3 + \frac{1}{\beta x})^{11}$ and $(\alpha x - \frac{1}{\beta x^3})^{11}$. We must equate the coefficient of $x^9$ in the first expansion with the coefficient of $x^{-9}$ in the second.

### The Hunt for $x^9$

To find the coefficient of any term, we utilize the general term formula:
$$T_{r+1} = \binom{n}{r} a^{n-r} b^r$$

For the first expansion, we set $a = \alpha x^3$ and $b = \frac{1}{\beta x}$. Substituting these into the formula, we obtain:
$$T_{r+1} = \binom{11}{r} (\alpha x^3)^{11-r} \left( \frac{1}{\beta x} \right)^r$$

Isolating the powers of $x$, we have $x^{3(11-r)}$ from the first part and $x^{-r}$ from the second. Combining these yields $x^{33-4r}$. To find the coefficient of $x^9$, we set the exponent equal to $9$:
$$33 - 4r = 9 \Rightarrow 4r = 24 \Rightarrow r = 6$$

Substituting $r=6$ back into the expression, the coefficient of $x^9$ is:
$$\binom{11}{6} \alpha^5 \beta^{-6}$$

### The Negative Sign Trap

Now, we examine the second expansion: $(\alpha x - \frac{1}{\beta x^3})^{11}$. Here, $a = \alpha x$ and $b = -\frac{1}{\beta x^3}$.

Using a new index $k$, the general term is:
$$T_{k+1} = \binom{11}{k} (\alpha x)^{11-k} \left( -\frac{1}{\beta x^3} \right)^k$$

Simplifying this expression, we get:
$$\binom{11}{k} \alpha^{11-k} (-1)^k \beta^{-k} x^{11-k} x^{-3k} = \binom{11}{k} \alpha^{11-k} (-1)^k \beta^{-k} x^{11-4k}$$

Setting the power of $x$ to $-9$:
$$11 - 4k = -9 \Rightarrow 4k = 20 \Rightarrow k = 5$$

The resulting coefficient is:
$$\binom{11}{5} \alpha^6 (-1)^5 \beta^{-5} = -\binom{11}{5} \alpha^6 \beta^{-5}$$

### The Beautiful Cancellation

We are given that these two coefficients are equal. Therefore:
$$\binom{11}{6} \alpha^5 \beta^{-6} = -\binom{11}{5} \alpha^6 \beta^{-5}$$

Since $\binom{11}{6} = \binom{11}{5}$, we can cancel these binomial coefficients from both sides. This leaves us with:
$$\alpha^5 \beta^{-6} = -\alpha^6 \beta^{-5}$$

Dividing both sides by $\alpha^5$ and multiplying by $\beta^6$, we find:
$$1 = -\alpha \beta \Rightarrow \alpha \beta = -1$$

### Final Calculation

The problem asks for the value of $(\alpha\beta)^2$. Squaring our result:
$$(\alpha\beta)^2 = (-1)^2 = 1$$

The final answer is **1**.

Animated Solution for Mathematics - Binomial Theorem: If the co-efficient of $x^{9}$ in $(α x^{3} + \frac{1}{β x})^{11}$ and the co-efficient of $x^{- 9}$ in $(α x - \frac{1}{β x ^{3}})^{11}$ are equal, then $(α β)^{2}$ is equal to ______.

Enter Numerical Value:

Visualized Solution

Problem Statement

General Term Formula

Simplifying Powers of $x$

Finding $r$ for $x^{9}$

Coefficient of $x^{9}$

General Term of the Second Expansion

Simplifying Powers of $x$ (Expansion 2)

Finding $k$ for $x^{- 9}$

Coefficient of $x^{- 9}$

Equating the Coefficients

Solving for $α β$

Final Answer Calculation

The Sigma Insight: General Term and Middle Term

Analyzing the Setup

The Hunt for $x^{9}$

The Negative Sign Trap

The Beautiful Cancellation

Final Calculation

Similar Questions

If the coefficients of $x^{7}$ in $(x^{2} + \frac{1}{b x})^{11}$ and $x^{- 7}$ in $(x - \frac{1}{b x ^{2}})^{11}$ , $b \neq = 0$ , are equal, then the value of $b$ is equal to:

If the coefficient of $x^{7}$ in $(a x^{2} - \frac{1}{b x})^{13}$ and the coefficient of $x^{- 5}$ in $(a x + \frac{1}{b x ^{2}})^{13}$ are equal, then $a^{4} b^{4}$ is equal to:

If the sum of the coefficients of all the positive even powers of $x$ in the binomial expansion of $(2 x^{3} + \frac{3}{x})^{10}$ is $5^{10} - β \cdot 3^{9}$ , then $β$ is equal to ____.

If the constant term in the binomial expansion of $(\frac{x ^{5/2}}{2} - \frac{4}{x ^{l}})^{9}$ is $- 84$ and the Coefficient of $x^{- 3 l}$ is $2^{α} β$ , where $β < 0$ is an odd number, Then $∣ α l - β ∣$ is equal to ______.

If the co-efficient of $x^{7}$ and $x^{8}$ in the expansion of $(2 + \frac{x}{3})^{n}$ are equal, then the value of $n$ is equal to ___

Let $a$ and $b$ be two nonzero real numbers. If the coefficient of $x^{5}$ in the expansion of $(a x^{2} + \frac{70}{27 b x})^{4}$ is equal to the coefficient of $x^{- 5}$ in the expansion of $(a x - \frac{1}{b x ^{2}})^{7}$ , then the value of $2 b$ is

The coefficient of $x^{4}$ in $(\frac{x}{2} - \frac{3}{x ^{2}})^{10}$ is

The Coefficient of $x^{- 6}$ in the expansion of $(\frac{4 x}{5} + \frac{5}{2 x ^{2}})^{9}$ is ______.

The absolute difference of the coefficients of $x^{10}$ and $x^{7}$ in the expansion of $(2 x^{2} + \frac{1}{2 x})^{11}$ is equal to

The coefficient of the middle term in the binomial expansion in powers of $x$ of $(1 + α x)^{4}$ and of $(1 - α x)^{6}$ is the same if $α$ equals

Enter Numerical Value:

Visualized Solution

Problem Statement

General Term Formula

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

Finding .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } r for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x9

Coefficient of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x9

General Term of the Second Expansion

Simplifying Powers of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x (Expansion 2)

Finding .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } k for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x−9

Coefficient of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x−9