Tr+1=9Cr(23x2)9−r(−3x1)r
=9Cr(23)9−r(−31)rx18−3r
For the term independent of x, put r=6
⇒T7=9C6(23)3(−31)6=9C3(61)3=3×2×19×8×7(61)3=(187)
⇒k=T7⇒k=187⇒18k=7
If the term independent of x in the expansion of (23x2−3x1)9 is k, then 18k is equal to:
Held on 3 Sept 2020 · Verified 6 Jul 2026.
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