JEE Mains · Maths · STD 11 - 7. binomial theoram
If the constant term in the expansion of \(\left(1+2 x-3 x^3\right)\left(\frac{3}{2} x^2-\frac{1}{3 x}\right)^9\) is \(p\), then \(108\) p is equal to ....................
- A \(43\)
- B \(54\)
- C \(77\)
- D \(55\)
Answer & Solution
Correct Answer
(B) \(54\)
Step-by-step Solution
Detailed explanation
\(\left(1+2 x-3 x^3\right)\left(\frac{3}{2} x^2-\frac{1}{3 x}\right)^9\) General term \(\mathrm{m}\left(\frac{3}{2} \mathrm{x}^2-\frac{1}{3 \mathrm{x}}\right)^9\)…
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