Here expansion given as :
(1+x+2x3)(3x22−13x)9
Now, to find the term undependent of x we have
(3x22−13x)9=r=0∑i=0 9Cr(3x22)9−r(−13x)r
=i=9∑i=0 9Cr(32)9−rx(2)(9−r)(−13)r(x)−r
=i=9∑i=0 9Cr×2r−9x18−3r .......... (1)
∴(1) [term independent of x in (1)]+(1) [Term containing x−1]+2 [Term containing x^{-3}$]
=(1)( 9Cr(−1)639−12×2−3)+(1)(0)+[2( 9C7)(−1)]
=(1)9!6!3!×127×18−2×9!7!2!×135×14
=7×8×96×127×18−2×8×92×135×14
=78−227
=21−454=1754
Term independent of x is 1754