find-the-greatest-coeeficient-of-1-x-2-2-i-have-applied-Q-125697-but-i-got-two-negative-values-of-n-after-i-have-made-k-2n-for-positive-coefficient-I-believe-it-should-have-a-greatest-coeefi

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Question Number 125881 by aurpeyz last updated on 14/Dec/20

$$findthegreatestcoeeficientof$$$${(1+\frac{x}{2})}^{-2}$$$$ihaveappliedQ.125697butigottwo$$$$negativevaluesofnafterihave$$$$madek=2nforpositivecoefficient.$$$$$$$$Ibelieveitshouldhaveagreatest$$$$coeeficientsince{\left(\frac{1}{2}\right)}^{k}decreasesas$$$$askincreases$$$$thanks$$$$whatdoido?$$

Commented by mr W last updated on 14/Dec/20

$$yes,thereisamaximum(aswellas$$$$minimum)coefficient.$$$${(1+\frac{x}{2})}^{-2}=\underset{k=0}{\sum ^{\mathrm{\infty }}}{(-1)}^k{C}_1^{k+1}\frac{1}{2^k}{x}^k$$$$a_{2n}=\frac{1}{2^{2n}}{C}_1^{2n+1}=\frac{2n+1}{2^{2n}}$$$$a_{2n+1}=-\frac{1}{2^{2n+1}}{C}_1^{2n+2}=-\frac{n+1}{2^{2n}}$$$$$$$$\frac{2n+1}{2^{2n}}>\frac{2(n+1)+1}{2^{2(n+1)}}$$$$4(2n+1)>2n+3$$$$n>-\frac{1}{6}$$$$\Rightarrow n_{min}=0$$$$\Rightarrow a_{0}isthelargestcoefficient$$$$\Rightarrow a_0=1$$

Commented by aurpeyz last updated on 15/Dec/20

$$wow.thatsnice.thanksalot.$$$$icannowsolveallproblemsonit$$

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