General term of the binomial expansion of (2x3−3x21)5 is
Tr+1=5Cr(2x3)5−r(3x2−1)r
⇒Tr+1=Cr5(2)5−r(−31)rx15−5r
For coefficient of x5, we must have
15−5r=5⇒r=2
So, required coefficient is
=C25(2)3(91)=10×8×91=980
JEE Main 2023 — Mathematics Algebra
The coefficient of x5 in the expansion of (2x3−3x21)5 is
Held on 13 Apr 2023 · Verified 6 Jul 2026.
980
9
8
326
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