Given expression (1+x+2x3)(3x22−13x)9
consider (3x22−13x)9
tr+1=9Cr(3x22)r(−13x)9−r=9Cr(32)9−r(−13)rx18−3r
Hence ,the general term in expansion by
(1+x+2x3)(3x22−13x)9is:
9Cr(32)q−r(−13)rx18−3r+9Cr(32)9−r(−13)rx19−3r+2.9Cr(32)q−r+(−13)rx21−3r
For term independent of x, put
18-3r=0 and 21-3r=0
r=6 and r=7
∴2nd term is not independent of x. (ie.r=6)
∴ term independent of x is
9C6(32)9−6(−13)6+29/7(32)9−7(−13)7
=9×8×7×6!6!×3×2×123.33−2×9×8×7!7!×2×3222×137
=848.133−364−235=21−454=1754