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Question

Prove the there is no term containing x10 in the expansion of (x22x)18

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Solution

The general term in the given expansion is given by
Tr+1=(1)r×18Cr×(x2)18r×(2x)rTr+1=(1)r×18Cr×2r×x(363r)Let Tr+1 Contain x10 then,363r=103r=26r=263


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