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Question Number 36166 by Rio Mike last updated on 29/May/18

$$\mathrm{Find}\mathrm{the}\mathrm{middle}\mathrm{term}\mathrm{in}$$$$\mathrm{the}\mathrm{expansion}\mathrm{of}{({\mathrm{x}}^{}+\frac{3}{\mathrm{x}})}^9$$

Commented by abdo.msup.com last updated on 30/May/18

$$wehave{(x+\frac{3}{x})}^9=\sum _{k=0}^9{C}_9^k{x}^k{\left(\frac{3}{x}\right)}^{9-k}$$$$=\sum _{k=0}^9{C}_9^k{3}^{9-k}{x}^{k+k-9}$$$$=\sum _{k=0}^9{C}_9^k{3}^{9-k}{x}^{2k-9}themiddle$$$$termis{ax}^{5}sowemusthave$$$$2k-9=5\Rightarrow k=7andthecoeffcient$$$$is{3}^{9-7}{C}_9^7=9.\frac{9!}{7!2!}=\frac{9.9.8}{2}$$$$=81\times 4=324sothecoefficientis$$$$354x^5.$$$$thecoefficientis324x^5.$$

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