JEE Mains · Maths · STD 11 - 7. binomial theoram
In the expansion of \(\left(9x-\dfrac{1}{3\sqrt{x}}\right)^{18}\), \(x>0\), if the term independent of \(x\) is \((221)k\), then \(k\) is equal to:
- A \(84\)
- B \(78\)
- C \(168\)
- D \(198\)
Answer & Solution
Correct Answer
(A) \(84\)
Step-by-step Solution
Detailed explanation
The general term in the expansion of \(\left(9x - \dfrac{1}{3\sqrt{x}}\right)^{18}\) is given by: \(T_{r+1} = ^{18}C_{r} (9x)^{18-r} \left(-\dfrac{1}{3\sqrt{x}}\right)^r\) \(T_{r+1} = ^{18}C_{r} 9^{18-r} \left(-\dfrac{1}{3}\right)^r x^{18-r} x^{-r/2}\)…
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