JEE Mains · Maths · STD 11 - 7. binomial theoram
The natural number \(m\), for which the coefficient of \(x\) in the binomial expansion of \(\left( x ^{ m }+\frac{1}{ x ^{2}}\right)^{22}\) is \(1540,\) is
- A \(19\)
- B \(3\)
- C \(13\)
- D \(18\)
Answer & Solution
Correct Answer
(C) \(13\)
Step-by-step Solution
Detailed explanation
\(T_{ r +1}={ }^{22} C _{ r }\left( x ^{ m }\right)^{22- r }\left(\frac{1}{ x ^{2}}\right)^{ r }=22 C _{ r } x ^{22 m - mr -2 r }\) \(=22 C _{ r } x\) \(\because 22 C _{3}=22 C _{19}=1540\) \(\therefore r =3\) or 19 \(Z 2 m - mr -2 r =1\) \(m =\frac{2 r +1}{22-5}\)…
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