Binomial Theorem
General and Middle Terms
Grade 11
Question:
<p>If the coefficients of \(x^{-2}\) and \(x^{-4}\) in the expansion of \(\left(x^{1/3} + \dfrac{1}{2x^{1/3}}\right)^{18}\), \((x > 0)\), are \(m\) and \(n\), respectively, then \(\dfrac{m}{n}\) is equal to</p>
<p>27</p>
<p>182</p>
<p>\(\dfrac{5}{4}\)</p>
<p>\(\dfrac{4}{5}\)</p>
Step-by-Step Solution
Key Concept: Use the binomial theorem to find the general term, then set the power of x equal to -2 and -4 separately to find which terms contribute these powers, and identify the corresponding binomial coefficients.
<p><strong>Step 1:</strong> Write the general term in the binomial expansion of $\left(x^{1/3} + \frac{1}{2x^{1/3}}\right)^{18}$:</p><p>$T_{r+1} = \binom{18}{r}(x^{1/3})^{18-r}\left(\frac{1}{2x^{1/3}}\right)^{r} = \binom{18}{r}\frac{1}{2^r}x^{\frac{18-r}{3}-\frac{r}{3}} = \binom{18}{r}\frac{1}{2^r}x^{\frac{18-2r}{3}}$</p><p><strong>Step 2:</strong> For the coefficient of $x^{-2}$, set the exponent equal to $-2$:</p><p>$\frac{18-2r}{3} = -2 \Rightarrow 18-2r = -6 \Rightarrow 2r = 24 \Rightarrow r = 12$</p><p>$m = \binom{18}{12}\frac{1}{2^{12}} = \binom{18}{12}\frac{1}{4096}$</p><p><strong>Step 3:</strong> For the coefficient of $x^{-4}$, set the exponent equal to $-4$:</p><p>$\frac{18-2r}{3} = -4 \Rightarrow 18-2r = -12 \Rightarrow 2r = 30 \Rightarrow r = 15$</p><p>$n = \binom{18}{15}\frac{1}{2^{15}}$</p><p><strong>Step 4:</strong> Calculate the ratio:</p><p>$\frac{m}{n} = \frac{\binom{18}{12}\frac{1}{2^{12}}}{\binom{18}{15}\frac{1}{2^{15}}} = \frac{\binom{18}{12}}{\binom{18}{15}} \cdot 2^{3}$</p><p>Since $\binom{18}{12} = \binom{18}{6}$ and $\binom{18}{15} = \binom{18}{3}$:</p><p>$\frac{m}{n} = \frac{\binom{18}{6}}{\binom{18}{3}} \cdot 8 = \frac{18564}{816} \cdot 8 = \frac{22.75 \cdot 816}{816} \cdot 8 = 91$</p><p>∴ Answer: <strong>91</strong></p>
Correct Answer: A