Question
Easy
The term independent of x in the expansion of $(x^{\frac{1}{3}}+\frac{1}{2x^{1/3}})^{18}, x>0$ is:
1
${}^{18}C_{9}\cdot\frac{1}{2^{9}}$
2
${}^{18}C_{7}\cdot\frac{1}{2^{7}}$
3
${}^{18}C_{5}\cdot\frac{1}{2^{5}}$
4
${}^{18}C_{3}\cdot\frac{1}{2^{3}}$
Question Details
Time to Solve: 12
Exam: HTET
Level/Paper: Level 3
Chapter: Algebra
Topic: Binomial Theorem
Correct Answer
Option A
Explanation
To determine the term independent of \( x \) in the expansion of \((x^{\frac{1}{3}}+\frac{1}{2x^{1/3}})^{18}\), we need to find the term where the power of \( x \) is zero. The general term in the binomial expansion of \((x^{\frac{1}{3}}+\frac{1}{2x^{1/3}})^{18}\) is given by: \[ T_k = \binom{18}{k} \left(x^{\frac{1}{3}}\right)^{18-k} \left(\frac{1}{2x^{1/3}}\right)^k \] Simplifying the expression, we have: \[ T_k = \binom{18}{k} \cdot x^{\frac{18-k}{3}} \cdot \frac{1}{2^k} \cdot x^{-\frac{k}{3}} \] \[ = \binom{18}{k} \cdot \frac{1}{2^k} \cdotтАжRead More
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