JEE Mains · Maths · STD 11 - 7. binomial theoram
Let \(\alpha > 0\), be the smallest number such that the expansion of \(\left(x^{\frac{2}{3}}+\frac{2}{x^3}\right)^{30}\) has a term \(\beta x^{-\alpha}, \beta \in N\). Then \(\alpha\) is equal to \(.............\).
- A \(2\)
- B \(4\)
- C \(6\)
- D \(8\)
Answer & Solution
Correct Answer
(A) \(2\)
Step-by-step Solution
Detailed explanation
\(T _{ r +1}={ }^{30} C _{ r }\left( x ^{2 / 3}\right)^{30- r }\left(\frac{2}{ x ^3}\right)^{ r }\) \(={ }^{30} C _{ r } \cdot 2^{ r } \cdot x ^{\frac{60-11 r }{3}}\) \(\frac{60-11 r }{3} < 0 \Rightarrow 11 r > 60 \Rightarrow r >\frac{60}{11} \Rightarrow r =6\)…
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