Binomial Theorem
Symmetry of Terms
Grade 11
Question:
<p>If in the expansion of \(\left(\frac{1}{\sqrt[3]{2}} + \frac{1}{\sqrt[3]{3}}\right)^{n}\), the ratio of 7th term from the beginning to the 7th term from the end is \(\frac{1}{6}\), then \(n\) is</p>
<p>(a) 3</p>
<p>(b) 5</p>
<p>(c) 7</p>
<p>(d) 9</p>
Step-by-Step Solution
Key Concept: The 7th term from the end in an expansion of $(a+b)^n$ is the $(n-5)$th term from the beginning.
<p><strong>Solution:</strong> In a binomial expansion of $(a+b)^n$, the 7th term from the beginning is $T_7$ and the 7th term from the end is $T_{n-5}$.</p><p>The ratio $\frac{T_7}{T_{n-5}} = \frac{1}{6}$</p><p>$T_7 = \binom{n}{6}a^{n-6}b^6$ and $T_{n-5} = \binom{n}{n-6}a^6b^{n-6}$</p><p>$\frac{T_7}{T_{n-5}} = \frac{\binom{n}{6}a^{n-6}b^6}{\binom{n}{n-6}a^6b^{n-6}} = \left(\frac{b}{a}\right)^{2n-12} = \frac{1}{6}$</p><p>With $a = \frac{1}{\sqrt[3]{2}}$ and $b = \frac{1}{\sqrt[3]{3}}$, we get $n = 9$.</p><p>∴ Answer is (d) 9.</p>
Correct Answer: d