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Question Number 157230 by Jamshidbek last updated on 21/Oct/21

(3x+1)^(100) Find this max Koeffitcient

$$\left(\mathrm{3x}+\mathrm{1}\right)^{\mathrm{100}} \:\: \\ $$$$\mathrm{Find}\:\mathrm{this}\:\mathrm{max}\:\mathrm{Koeffitcient} \\ $$

Answered by mr W last updated on 21/Oct/21

a_n =3^n C_n ^(100) a_(n+1) =3^(n+1) C_(n+1) ^(100) (a_(n+1) /a_n )=((3C_(n+1) ^(100) )/C_n ^(100) )=((3(100āˆ’n))/((n+1)))<1 299<4n n>((299)/4)ā‡’nā‰„75 maximum is at n=75 max=a_(75) =3^(75) C_(75) ^(100)

$${a}_{{n}} =\mathrm{3}^{{n}} {C}_{{n}} ^{\mathrm{100}} \\ $$$${a}_{{n}+\mathrm{1}} =\mathrm{3}^{{n}+\mathrm{1}} {C}_{{n}+\mathrm{1}} ^{\mathrm{100}} \\ $$$$\frac{{a}_{{n}+\mathrm{1}} }{{a}_{{n}} }=\frac{\mathrm{3}{C}_{{n}+\mathrm{1}} ^{\mathrm{100}} }{{C}_{{n}} ^{\mathrm{100}} }=\frac{\mathrm{3}\left(\mathrm{100}āˆ’{n}\right)}{\left({n}+\mathrm{1}\right)}<\mathrm{1} \\ $$$$\mathrm{299}<\mathrm{4}{n} \\ $$$${n}>\frac{\mathrm{299}}{\mathrm{4}}\Rightarrow{n}\geqslant\mathrm{75} \\ $$$${maximum}\:{is}\:{at}\:{n}=\mathrm{75} \\ $$$${max}={a}_{\mathrm{75}} =\mathrm{3}^{\mathrm{75}} {C}_{\mathrm{75}} ^{\mathrm{100}} \\ $$

Commented by Tawa11 last updated on 21/Oct/21

Great sir

$$\mathrm{Great}\:\mathrm{sir} \\ $$

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