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JEE Mains · Maths · STD 11 - 7. binomial theoram

प्राकृत संख्या \(m\), जिसके लिए \(\left( x ^{ m }+\frac{1}{ x ^{2}}\right)^{22}\) के द्विपद प्रसार में \(x\) का गुणांक \(1540\) है

  1. A \(19\)
  2. B \(3\)
  3. C \(13\)
  4. D \(18\)
Verified Solution

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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