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Question

Find the term independent of x in the expansion of (1+x+2x3)(32x213x)9


Solution

In the expansion of E=(32x213x)9, we have

Tr+1=(1)r.9Cr.(32x2)(9r)(13x)r

Tr+1=(1)r.9Cr.3(92r)2(9r).x(183r).

(1+x+2x3)[(a0×1x3+a1×1x+a2)from E]

=(1+x+2x3)[{(1)7.9C7.e522×1x3}+{(1)6.9C6.3323×x0}]

x=1183r=1r is fraction183r=0r=6 and 183r=3r=7

=(1+x+2x3)[127x3+718]

required term =(227+718)=1754

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