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JEE Main 20205 SeptemberMorning Shift
Exam
JEE Main
Year
2020
Date
5 September
Shift
Morning Shift
Subject
Mathematics
Chapter
Binomial Theorem

JEE Main 2020 · 5 September · Morning Shift · Mathematics · Binomial Theorem

The natural number m , for which the coefficient of x in the...

The natural number mm, for which the coefficient of xx in the binomial expansion of ${\left({x}^{m} + \dfrac{1}{{x}^{2}}\right)}^{22} is 1540, is

Correct answer

13

Step-by-step solution

The general term in the binomial expansion,

Tr+1=Cr22(xm)22−rx−2r{T}_{r + 1} = Cr 22 {\left({x}^{m}\right)}^{22 - r}{x}^{- 2r}

⇒Tr+1=Cr22xm(22−r)−2r⇒ {T}_{r + 1} = Cr 22 {x}^{m ( 22 - r ) - 2r}

For the coefficient of xx

22m−mr−2r=122m - mr - 2r = 1

⇒22m−1=r(m+2)⇒ 22m - 1 = r ( m + 2 )

⇒r=22m−1m+2⇒ r = \dfrac{22m - 1}{m + 2}

⇒r=22m+44−45m+2⇒ r = \dfrac{22m + 44 - 45}{m + 2}

r=22−3.3.5m+2⇒22−r=3⋅3⋅5m+2r = 22 - \dfrac{3 . 3 . 5}{m + 2} ⇒ 22 - r = \dfrac{3 \cdot 3 \cdot 5}{m + 2}

Since, Cr22=C22−r22Cr 22 = C22 - r 22

So possible values of m=1,3,7,13,43m = 1, 3, 7, 13, 43, but C322=1540⇒22−r=3⇒m=13C3 22 = 1540 ⇒ 22 - r = 3 ⇒ m = 13

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