Verified Solution • 2025–26 Session moderate

QJEEM20Ma3aff: JEE Main 2020 Mathematics - Class 11 JEEM20Ma3aff Solution: The natural number m, for which the coefficient of x in the…

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

(xm+1x2)22{\left( {{x^m} + {1 \over {{x^2}}}} \right)^{22}} is 1540, is .............

Step-by-Step Verified Solution

General term,

Tr+1=22Cr(xm)22−r(1x2)r=22Crx22m−mr−2r{T_{r + 1}} = {}^{22}{C_r}{({x^m})^{22 - r}}{\left( {{1 \over {{x^2}}}} \right)^r} = {}^{22}{C_r}{x^{22m - mr - 2r}}

∵\because 22C3=22C19=1540{}^{22}{C_3} = {}^{22}{C_{19}} = 1540

∴\therefore r=3 or 19r = 3\,or\,19

22m−mr−2r=122m - mr - 2r = 1

m=2r+122−5m = {{2r + 1} \over {22 - 5}}

When r=3r = 3 , m=719∉Nm = {7 \over {19}} \notin N

When r=19r = 19 , m=38+122−19=393=13m = {{38 + 1} \over {22 - 19}} = {{39} \over 3} = 13

∴\therefore m=13m = 13

Related Learning Resources & Exam Preparation