The rth term of (2x51−x511)15
⇒Tr+1=Cr15(2x51)15−r(x51−1)r
For term having x−1
515−r−5r=−1⇒r=10 ⇒m=C1015⋅25⋅(−1)10
For term having x−3
515−2r=−3 ⇒r=15 ⇒n=C1515⋅20⋅(−1)15
∴mn2=C1015⋅25=C515⋅25
On comparing with m⋅n2=Cr15⋅2r, we get r=5
JEE Main 2022 — Mathematics Algebra
Let the coefficients of x−1 and x−3 in the expansion of (2x51−x511)15,x>0, be mand n respectively. If r is a positive integer such mn2=Cr.152r, then the value of r is equal to ______.
Held on 29 Jun 2022 · Verified 6 Jul 2026.
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