Understanding the Binomial Theorem and General Term

The foundation for solving this problem lies in the Binomial Theorem, which provides a systematic way to expand expressions of the form $(p+q)^n$. A crucial part of this theorem is the general term, often denoted as $T_{r+1}$, which allows us to find any specific term in the expansion without writing out the entire series.

The formula for the general term of $(p+q)^n$ is: $T_{r+1} = \binom{n}{r} p^{n-r} q^r$ Where:

• $\binom{n}{r}$ (read as "n choose r") is the binomial coefficient, calculated as $\frac{n!}{r!(n-r)!}$.
• $p$ is the first term of the binomial.
• $q$ is the second term of the binomial.
• $n$ is the power to which the binomial is raised.
• $r$ is the index of the term (starting from $r=0$ for the first term, $r=1$ for the second, and so on).

To find the coefficient of a specific power of $x$ (say, $x^k$) in a binomial expansion, we follow these general steps:

1. Identify Components: Determine $p$, $q$, and $n$ from the given binomial expression.
2. Formulate General Term: Write down the general term $T_{r+1}$ using the formula above.
3. Isolate $x$ Power: Collect all $x$ terms in the general term and simplify their exponents to express the power of $x$ in terms of $r$.
4. Solve for $r$: Equate the simplified power of $x$ to the desired power $k$ and solve for the integer value of $r$.
5. Extract Coefficient: Substitute the obtained value of $r$ back into the general term, excluding any $x$ terms, to find the specific coefficient.

Analyzing the First Expression: $\left( {a{x^2} + {1 \over {2bx}}} \right)^{11}$

Let's apply the steps to the first expression: Given binomial: $\left( {a{x^2} + {1 \over {2bx}}} \right)^{11}$ Here, we identify:

• $p = ax^2$
• $q = \frac{1}{2bx}$
• $n = 11$

Step 1: Formulate the General Term $T_{r+1}$ Using the binomial general term formula: $T_{r+1} = \binom{11}{r} (ax^2)^{11-r} \left(\frac{1}{2bx}\right)^r$

Step 2: Separate Constants and $x$ Terms To clearly see the power of $x$, we distribute the exponents and separate the components: $T_{r+1} = \binom{11}{r} a^{11-r} (x^2)^{11-r} \frac{1}{(2b)^r} \frac{1}{x^r}$ Applying exponent rules $(x^m)^n = x^{mn}$ and $\frac{1}{x^k} = x^{-k}$: $T_{r+1} = \binom{11}{r} a^{11-r} x^{2(11-r)} (2b)^{-r} x^{-r}$

Step 3: Combine $x$ Terms and Simplify Exponent Now, combine the $x$ terms using $x^m \cdot x^n = x^{m+n}$: $T_{r+1} = \binom{11}{r} a^{11-r} (2b)^{-r} x^{2(11-r) - r}$ $T_{r+1} = \binom{11}{r} a^{11-r} (2b)^{-r} x^{22-2r - r}$ $T_{r+1} = \binom{11}{r} a^{11-r} (2b)^{-r} x^{22-3r}$

Step 4: Solve for $r$ for the Coefficient of $x^7$ We are looking for the coefficient of $x^7$, so we set the power of $x$ equal to $7$: $22-3r = 7$ Subtract $22$ from both sides: $-3r = 7 - 22$ $-3r = -15$ Divide by $-3$: $r = 5$

**Step 5: Extract the Coefficient $C_1$} Substitute $r=5$ back into the coefficient part of $T_{r+1}$ (everything except $x^{22-3r}$): $C_1 = \binom{11}{5} a^{11-5} (2b)^{-5}$ $C_1 = \binom{11}{5} a^6 \frac{1}{(2b)^5}$ $C_1 = \binom{11}{5} a^6 \frac{1}{32b^5}$

Tip: Always double-check your exponent calculations, especially when dealing with nested exponents or terms in the denominator. A small error here can lead to an incorrect value of $r$ or an incorrect coefficient.

Analyzing the Second Expression: $\left( {ax - {1 \over {3b{x^2}}}} \right)^{11}$

Now, let's process the second expression: Given binomial: $\left( {ax - {1 \over {3b{x^2}}}} \right)^{11}$ Here, we identify:

• $p = ax$
• $q = -\frac{1}{3bx^2}$ (Crucially, remember the negative sign with $q$)
• $n = 11$

Step 1: Formulate the General Term $T_{r+1}$ $T_{r+1} = \binom{11}{r} (ax)^{11-r} \left(-\frac{1}{3bx^2}\right)^r$

Step 2: Separate Constants and $x$ Terms $T_{r+1} = \binom{11}{r} a^{11-r} x^{11-r} (-1)^r \frac{1}{(3b)^r} \frac{1}{(x^2)^r}$ $T_{r+1} = \binom{11}{r} a^{11-r} x^{11-r} (-1)^r (3b)^{-r} x^{-2r}$

Step 3: Combine $x$ Terms and Simplify Exponent $T_{r+1} = \binom{11}{r} a^{11-r} (-1)^r (3b)^{-r} x^{11-r - 2r}$ $T_{r+1} = \binom{11}{r} a^{11-r} (-1)^r (3b)^{-r} x^{11-3r}$

Step 4: Solve for $r$ for the Coefficient of $x^{-7}$ We are looking for the coefficient of $x^{-7}$, so we set the power of $x$ equal to $-7$: $11-3r = -7$ Add $3r$ to both sides and add $7$ to both sides: $11+7 = 3r$ $18 = 3r$ $r = 6$

**Step 5: Extract the Coefficient $C_2$} Substitute $r=6$ back into the coefficient part of $T_{r+1}$: $C_2 = \binom{11}{6} a^{11-6} (-1)^6 (3b)^{-6}$ Since $(-1)^6 = 1$ (any even power of $-1$ is $1$): $C_2 = \binom{11}{6} a^5 \frac{1}{(3b)^6}$ $C_2 = \binom{11}{6} a^5 \frac{1}{729b^6}$

Common Mistake: A very common error is to overlook the negative sign in the second term of the binomial, especially when raised to a power $r$. If $r$ were odd, the coefficient would have an additional negative sign. In this case, $r=6$ (even), so $(-1)^6=1$.

**Equating the Coefficients and Solving for $ab$}

The problem states that the coefficient of $x^7$ from the first expression ($C_1$) is equal to the coefficient of $x^{-7}$ from the second expression ($C_2$). $C_1 = C_2$ $\binom{11}{5} a^6 \frac{1}{32b^5} = \binom{11}{6} a^5 \frac{1}{729b^6}$

Key Simplification: We know a property of binomial coefficients that $\binom{n}{r} = \binom{n}{n-r}$. In our case, $\binom{11}{5} = \binom{11}{11-5} = \binom{11}{6}$. Since these binomial coefficients are equal, we can cancel them from both sides of the equation: $a^6 \frac{1}{32b^5} = a^5 \frac{1}{729b^6}$

Now, we perform algebraic manipulation to isolate the product $ab$: Divide both sides by $a^5$ (assuming $a \neq 0$, as coefficients would be zero otherwise): $a \cdot \frac{1}{32b^5} = \frac{1}{729b^6}$ Multiply both sides by $32b^5$: $a = \frac{32b^5}{729b^6}$ Simplify the $b$ terms ($b^5/b^6 = 1/b$, assuming $b \neq 0$, as terms would be undefined otherwise): $a = \frac{32}{729b}$ Finally, multiply both sides by $b$: $ab = \frac{32}{729}$ To remove the denominator, multiply both sides by $729$: $729ab = 32$

Key Takeaway: Recognizing and utilizing the symmetry property of binomial coefficients ($\binom{n}{r} = \binom{n}{n-r}$) can significantly simplify calculations in such problems. Careful handling of signs and exponents when setting up and solving for $r$ is paramount. This problem highlights a standard approach to finding specific terms and their coefficients in binomial expansions.

The final derived relationship is $\mathbf{729ab = 32}$.