JEE Mains · Maths · STD 11 - 7. binomial theoram
The constant term in the expansion of \(\left(2 x+\frac{1}{x^7}+3 x^2\right)^5 \text { is }........\).
- A \(1089\)
- B \(1080\)
- C \(1050\)
- D \(1562\)
Answer & Solution
Correct Answer
(B) \(1080\)
Step-by-step Solution
Detailed explanation
General term is \(\sum \frac{5 !(2 x)^{n_1}\left(x^{-7}\right)^{n_2}\left(3 x^2\right)^{n_3}}{n_{1} ! n_{2} ! n_{3} !}\) For constant term, \(n _1+2 n _3=7 n _2\) \(n _1+ n _2+ n _3=5\) Only possibility \(n _1=1, n _2=1, n _3=3\) \(\Rightarrow\) constant term \(=1080\)
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