Detailed Solution: Finding Coefficients in Binomial Expansions

1. Key Concept: The General Term in Binomial Expansion

The Binomial Theorem states that for any real numbers $A$ and $B$, and any non-negative integer $n$, the expansion of $(A+B)^n$ is given by: $(A+B)^n = \sum_{r=0}^n \binom{n}{r} A^{n-r} B^r$ The $(r+1)^{th}$ term in this expansion, denoted as $T_{r+1}$, is: $T_{r+1} = \binom{n}{r} A^{n-r} B^r$ This formula is fundamental for finding a specific term or the coefficient of a particular power of $x$ without performing the entire expansion.

2. Analyzing the First Expression: Coefficient of $x^7$ in ${\left( {ax - {1 \over {b{x^2}}}} \right)^{13}}$

Step 2.1: Identify the components of the binomial. For the expression ${\left( {ax - {1 \over {b{x^2}}}} \right)^{13}}$:

• $A = ax$ (the first term)
• $B = -{1 \over {b{x^2}}}$ (the second term, including its sign)
• $n = 13$ (the exponent of the binomial)

Step 2.2: Write the general term $T_{r+1}$. Substitute these values into the general term formula: $T_{r+1} = \binom{13}{r} (ax)^{13-r} \left(-\frac{1}{bx^2}\right)^r$ Now, we simplify this expression by separating the constant terms, $a$, $b$, and $x$ terms. $T_{r+1} = \binom{13}{r} a^{13-r} x^{13-r} \left(-\frac{1}{b}\right)^r \left(\frac{1}{x^2}\right)^r$ $T_{r+1} = \binom{13}{r} a^{13-r} x^{13-r} \left(-\frac{1}{b}\right)^r x^{-2r}$ Combine the powers of $x$: $T_{r+1} = \binom{13}{r} a^{13-r} \left(-\frac{1}{b}\right)^r x^{13-r-2r}$ $T_{r+1} = \binom{13}{r} a^{13-r} \left(-\frac{1}{b}\right)^r x^{13-3r}$ The coefficient of the $(r+1)^{th}$ term is therefore $\binom{13}{r} a^{13-r} \left(-\frac{1}{b}\right)^r$, and the power of $x$ is $13-3r$.

Step 2.3: Determine the value of $r$ for the coefficient of $x^7$. To find the coefficient of $x^7$, we set the exponent of $x$ in the general term equal to 7: $13 - 3r = 7$ Subtract 13 from both sides: $-3r = 7 - 13$ $-3r = -6$ Divide by -3: $r = 2$ Since $r=2$ is a non-negative integer and $0 \le r \le n$ (i.e., $0 \le 2 \le 13$), this is a valid value for $r$.

Step 2.4: Calculate the coefficient of $x^7$. Substitute $r=2$ into the coefficient part of the general term: $\text{Coefficient of } x^7 = \binom{13}{2} a^{13-2} \left(-\frac{1}{b}\right)^2$ $\text{Coefficient of } x^7 = \binom{13}{2} a^{11} \left(\frac{1}{b^2}\right)$ $\text{Coefficient of } x^7 = \binom{13}{2} \frac{a^{11}}{b^2}$

3. Analyzing the Second Expression: Coefficient of $x^{-5}$ in ${\left( {ax + {1 \over {b{x^2}}}} \right)^{13}}$

Step 3.1: Identify the components of the binomial. For the expression ${\left( {ax + {1 \over {b{x^2}}}} \right)^{13}}$:

• $A = ax$
• $B = {1 \over {b{x^2}}}$ (Note the positive sign here)
• $n = 13$

Step 3.2: Write the general term $T_{r+1}$. Substitute these values into the general term formula: $T_{r+1} = \binom{13}{r} (ax)^{13-r} \left(\frac{1}{bx^2}\right)^r$ Simplify as before: $T_{r+1} = \binom{13}{r} a^{13-r} x^{13-r} \left(\frac{1}{b}\right)^r \left(\frac{1}{x^2}\right)^r$ $T_{r+1} = \binom{13}{r} a^{13-r} \left(\frac{1}{b}\right)^r x^{13-r-2r}$ $T_{r+1} = \binom{13}{r} a^{13-r} \left(\frac{1}{b}\right)^r x^{13-3r}$ Notice that the power of $x$ is the same as in the first expansion, but the coefficient differs due to the sign of $B$.

Step 3.3: Determine the value of $r$ for the coefficient of $x^{-5}$. To find the coefficient of $x^{-5}$, we set the exponent of $x$ equal to -5: $13 - 3r = -5$ Subtract 13 from both sides: $-3r = -5 - 13$ $-3r = -18$ Divide by -3: $r = 6$ Again, $r=6$ is a valid non-negative integer ($0 \le 6 \le 13$).

Step 3.4: Calculate the coefficient of $x^{-5}$. Substitute $r=6$ into the coefficient part of the general term: $\text{Coefficient of } x^{-5} = \binom{13}{6} a^{13-6} \left(\frac{1}{b}\right)^6$ $\text{Coefficient of } x^{-5} = \binom{13}{6} a^7 \frac{1}{b^6}$ $\text{Coefficient of } x^{-5} = \binom{13}{6} \frac{a^7}{b^6}$

4. Equating the Coefficients and Solving for $a^4b^4$

The problem states that the coefficient of $x^7$ in the first expression and the coefficient of $x^{-5}$ in the second expression are equal. So, we set the two coefficients we found equal to each other: $\binom{13}{2} \frac{a^{11}}{b^2} = \binom{13}{6} \frac{a^7}{b^6}$ Our goal is to find the value of $a^4b^4$. We can rearrange the equation to isolate $a$ and $b$ terms on one side and the binomial coefficients on the other.

First, divide both sides by $a^7$: $\binom{13}{2} \frac{a^{11}}{b^2 a^7} = \binom{13}{6} \frac{1}{b^6}$ $\binom{13}{2} \frac{a^4}{b^2} = \binom{13}{6} \frac{1}{b^6}$ Now, multiply both sides by $b^6$: $\binom{13}{2} a^4 \frac{b^6}{b^2} = \binom{13}{6}$ $\binom{13}{2} a^4 b^4 = \binom{13}{6}$ Finally, divide by $\binom{13}{2}$ to isolate $a^4 b^4$: $a^4 b^4 = \frac{\binom{13}{6}}{\binom{13}{2}}$ Now, we calculate the values of the binomial coefficients: $\binom{13}{2} = \frac{13!}{2!(13-2)!} = \frac{13 \times 12}{2 \times 1} = 13 \times 6 = 78$ $\binom{13}{6} = \frac{13!}{6!(13-6)!} = \frac{13 \times 12 \times 11 \times 10 \times 9 \times 8}{6 \times 5 \times 4 \times 3 \times 2 \times 1}$ Let's simplify $\binom{13}{6}$: $6 \times 2 = 12$, so cancel 12 in numerator. $5 \times 1 = 5$, so $10/5 = 2$. $4$ in denominator, and $8$ in numerator, so $8/4 = 2$. $3$ in denominator, and $9$ in numerator, so $9/3 = 3$. $\binom{13}{6} = 13 \times 1 \times 11 \times 2 \times 3 \times 2 = 13 \times 11 \times 12 = 13 \times 132 = 1716$ Now, substitute these values back into the equation for $a^4b^4$: $a^4 b^4 = \frac{1716}{78}$ $a^4 b^4 = 22$

5. Important Tips and Common Mistakes

• Sign Errors: Be extremely careful with negative signs in the binomial term, especially when raised to a power $r$. In the first expression, $(-1/bx^2)^r$ becomes $(-1)^r (1/b^r) (1/x^{2r})$. If $r$ is even, $(-1)^r=1$; if $r$ is odd, $(-1)^r=-1$. In this specific problem, $r=2$ (even) and $r=6$ (even), so the negative signs cancel out, but it's a crucial point to remember.
• Exponent of x: Make sure to correctly combine all powers of $x$ from both terms $A^{n-r}$ and $B^r$. For example, $(ax)^{13-r}$ contributes $x^{13-r}$ and $(1/x^2)^r$ contributes $x^{-2r}$.
• Value of $r$: The value of $r$ must always be a non-negative integer ($0, 1, 2, \dots, n$). If you get a fractional or negative value for $r$, it means that power of $x$ does not exist in the expansion.
• Simplification of Binomial Coefficients: When dividing binomial coefficients, it's often easier to write them out using factorials or the expanded form and then cancel common terms, rather than calculating each large number individually and then dividing.

6. Conclusion and Key Takeaway

This problem effectively tests the application of the general term formula in binomial expansions. By systematically identifying the components of each binomial, writing out the general term, isolating the power of $x$, and then solving for $r$, we can determine the specific coefficients. The final step involves equating these coefficients and algebraic manipulation to solve for the required expression, $a^4b^4$. Mastering the general term formula and careful algebraic handling are key to solving such problems efficiently and accurately.