Understanding the Binomial Theorem and General Term

The problem requires us to find the term independent of $x$ in a binomial expansion. For a binomial expression of the form $(a+b)^n$, the general term, often denoted as $T_{r+1}$ (the $(r+1)^{th}$ term), is given by the formula: $T_{r+1} = {}^n C_r a^{n-r} b^r$ where ${}^n C_r = \frac{n!}{r!(n-r)!}$ is the binomial coefficient. This formula allows us to find any specific term in the expansion without having to expand the entire expression.

Step-by-Step Solution

1. Identify $a, b,$ and $n$ Given the expansion: ${\left( {{3 \over 2}{x^2} - {1 \over {3x}}} \right)^9}$ We compare this with $(a+b)^n$:

• $a = {3 \over 2}{x^2}$
• $b = -{1 \over {3x}}$ (Note the negative sign is crucial)
• $n = 9$

2. Apply the General Term Formula Substitute these values into the general term formula: $T_{r+1} = {}^9 C_r \left( {{3 \over 2}{x^2}} \right)^{9-r} \left( { - {1 \over {3x}}} \right)^r$

3. Separate Coefficients and Powers of $x$ To determine the term independent of $x$, we need to isolate all the $x$ terms. This step involves carefully applying exponent rules. $T_{r+1} = {}^9 C_r \left( {{3 \over 2}} \right)^{9-r} (x^2)^{9-r} \left( { - {1 \over 3}} \right)^r \left( {{1 \over x}} \right)^r$ Using the exponent rules $(x^m)^n = x^{mn}$ and $\frac{1}{x^p} = x^{-p}$: $T_{r+1} = {}^9 C_r \left( {{3 \over 2}} \right)^{9-r} x^{2(9-r)} \left( { - {1 \over 3}} \right)^r x^{-r}$ Now, combine the coefficients and the powers of $x$: $T_{r+1} = {}^9 C_r \left( {{3 \over 2}} \right)^{9-r} \left( { - {1 \over 3}} \right)^r \cdot x^{18-2r-r}$ $T_{r+1} = {}^9 C_r \left( {{3 \over 2}} \right)^{9-r} \left( { - {1 \over 3}} \right)^r \cdot x^{18-3r}$ This expression now clearly shows the numerical coefficient part and the $x$ dependent part.

4. Determine $r$ for the Term Independent of $x$ A term is "independent of $x$" if the power of $x$ in that term is zero. Therefore, we set the exponent of $x$ equal to zero: $18 - 3r = 0$ Solving for $r$: $3r = 18$ $r = 6$ This means the term independent of $x$ is the $(6+1)^{th}$, or $7^{th}$ term.

5. Calculate the Value of the Term (k) Now that we have found $r=6$, we substitute this value back into the coefficient part of our general term expression (excluding $x^{18-3r}$): $k = {}^9 C_6 \left( {{3 \over 2}} \right)^{9-6} \left( { - {1 \over 3}} \right)^6$ Calculate the binomial coefficient ${}^9 C_6$: Since ${}^n C_r = {}^n C_{n-r}$, we have ${}^9 C_6 = {}^9 C_{9-6} = {}^9 C_3$. ${}^9 C_3 = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = 3 \times 4 \times 7 = 84$ Now substitute this back and perform the arithmetic: $k = 84 \times \left( {{3 \over 2}} \right)^3 \times \left( { - {1 \over 3}} \right)^6$ $k = 84 \times \left( \frac{3^3}{2^3} \right) \times \left( \frac{(-1)^6}{3^6} \right)$ $k = 84 \times \frac{27}{8} \times \frac{1}{729}$ To simplify, notice that $27 = 3^3$ and $729 = 3^6$. $k = 84 \times \frac{27}{8 \times 27 \times 27}$ $k = 84 \times \frac{1}{8 \times 27}$ $k = \frac{84}{216}$ We can simplify this fraction. Divide both numerator and denominator by their greatest common divisor. Both are divisible by 12: $84 \div 12 = 7$ and $216 \div 12 = 18$. $k = \frac{7}{18}$ Let's cross-check with the provided solution's intermediate value for $k$: $\frac{21}{54}$. Indeed, $\frac{21}{54} = \frac{7 \times 3}{18 \times 3} = \frac{7}{18}$. So, the calculated value of $k$ is correct.

6. Calculate $18k$ The problem asks for the value of $18k$. $18k = 18 \times \left( \frac{7}{18} \right)$ $18k = 7$

Tips for Success & Common Mistakes to Avoid

• Sign Errors: Always pay close attention to negative signs within the terms of the binomial, as an odd power of a negative number results in a negative value, while an even power results in a positive value. Here, $(-1/3)^6$ correctly became $1/3^6$.
• Exponent Rules: Be meticulous when combining powers of $x$. A common mistake is to misapply $(x^m)^n = x^{mn}$ or $1/x^p = x^{-p}$.
• Simplification: Break down large calculations into smaller, manageable steps. Look for opportunities to cancel common factors before multiplying large numbers. For example, recognizing $27 = 3^3$ and $729 = 3^6$ simplified the fraction significantly.
• Understanding the Question: Ensure you correctly identify what the question is asking for (e.g., the value of $r$, the term itself, or a multiple of the term).

Summary This problem effectively demonstrates the application of the binomial theorem to find a specific term in an expansion. The process involves identifying the components $a$, $b$, and $n$, constructing the general term, isolating the powers of the variable ($x$), and then setting the exponent of the variable to zero to find the specific term number. Finally, substituting this term number back into the coefficient part yields the desired value. Mastery of exponent rules and careful arithmetic are crucial for success in these types of problems.