JEE Mains · Maths · STD 11 - 7. binomial theoram
If the term without \(x\) in the expansion of \(\left( x ^{\frac{2}{3}}+\frac{\alpha}{ x ^3}\right)^{22}\) is \(7315\) , then \(|\alpha|\) is equal to \(...........\).
- A \(2\)
- B \(1\)
- C \(4\)
- D \(6\)
Answer & Solution
Correct Answer
(B) \(1\)
Step-by-step Solution
Detailed explanation
\(T _{ r +1}={ }^{22} C _{ r } \cdot\left( x ^{\frac{2}{3}}\right)^{22- r } \cdot(\alpha)^{ r }, x ^{-3 r }\) \(={ }^{22} C _{ r } \cdot x ^{\frac{44}{3}-\frac{2 r }{3}-3 r }(\alpha)^{ r }\) \(\frac{44}{3}=\frac{11 r }{3}\) \(r =4\) \({ }^{22} C _4 \cdot \alpha^4=7315\)…
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