if-in-the-expansion-of-1-x-n-the-coefficient-of-x-9-is-the-aritmetic-mean-of-the-coeficients-of-x-8-and-x-10-find-the-possible-value-of-n-where-n-is-a-positive-integer-

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Question Number 87175 by john santu last updated on 03/Apr/20

$$\mathrm{if}\mathrm{in}\mathrm{the}\mathrm{expansion}\mathrm{of}{(1+\mathrm{x})}^{\mathrm{n}}\mathrm{the}$$$$\mathrm{coefficient}\mathrm{of}{\mathrm{x}}^9\mathrm{is}\mathrm{the}\mathrm{aritmetic}$$$$\mathrm{mean}\mathrm{of}\mathrm{the}\mathrm{coeficients}\mathrm{of}{\mathrm{x}}^8\mathrm{and}$$$${\mathrm{x}}^{10}.\mathrm{find}\mathrm{the}\mathrm{possible}\mathrm{value}$$$$\mathrm{of}\mathrm{n}\mathrm{where}\mathrm{n}\mathrm{is}\mathrm{a}\mathrm{positive}\mathrm{integer}$$

Commented by jagoll last updated on 03/Apr/20

$$\mathrm{i}\mathrm{can}\mathrm{try}$$$$2{\mathrm{C}}_9^{\mathrm{n}}={\mathrm{C}}_8^{\mathrm{n}}+{\mathrm{C}}_{10}^{\mathrm{n}}$$$$2\times \frac{\mathrm{n}!}{9!(\mathrm{n}-9)!}=\frac{\mathrm{n}!}{8!(\mathrm{n}-8)!}+\frac{\mathrm{n}!}{10!(\mathrm{n}-10)!}$$$$\frac{2}{9(\mathrm{n}-9)}=\frac{1}{1}+\frac{1}{10.9.(\mathrm{n}-10)(\mathrm{n}-9)}$$$$\frac{20(\mathrm{n}-10)-1}{10.9.(\mathrm{n}-9)(\mathrm{n}-10)}=1$$$$20\mathrm{n}-201=90({\mathrm{n}}^2-19\mathrm{n}+90)$$$${90\mathrm{n}}^2-1690\mathrm{n}+8301=0$$

Commented by john santu last updated on 03/Apr/20

$$\mathrm{good}\mathrm{sir}$$

Answered by john santu last updated on 03/Apr/20

$${2\mathrm{C}}_9^{\mathrm{n}}={\mathrm{C}}_8^{\mathrm{n}}+{\mathrm{C}}_{10}^{\mathrm{n}}$$$${4\mathrm{C}}_9^{\mathrm{n}}=({\mathrm{C}}_8^{\mathrm{n}}+{\mathrm{C}}_9^{\mathrm{n}})+({\mathrm{C}}_9^{\mathrm{n}}+{\mathrm{C}}_{10}^{\mathrm{n}})$$$${4\mathrm{C}}_9^{\mathrm{n}}={\mathrm{C}}_9^{\mathrm{n}+1}+{\mathrm{C}}_{10}^{\mathrm{n}+1}={\mathrm{C}}_{10}^{\mathrm{n}+2}$$$$\frac{4\mathrm{n}!}{9!(\mathrm{n}-9)!}=\frac{(\mathrm{n}+2)!}{10!(\mathrm{n}-8)!}$$$$4=\frac{{\mathrm{n}}^2+3\mathrm{n}+2}{10(\mathrm{n}-8)}\Rightarrow {\mathrm{n}}^2-37\mathrm{n}+322=0$$$$(\mathrm{n}-14)(\mathrm{n}-23)=0$$$$\mathrm{n}=14\mathrm{or}\mathrm{n}=23$$$$$$

Commented by peter frank last updated on 03/Apr/20

$$good$$

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