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Question Number 125697 by aurpeyz last updated on 13/Dec/20

$$whatisthelargestcoefficientof$$$${(4+3x)}^{-5}?$$

Answered by mr W last updated on 13/Dec/20

$${(4+3x)}^{-5}$$$$=4^{-5}{[1-(-\frac{3x}{4})]}^{-5}$$$$=\frac{1}{1024}\underset{k=0}{\sum ^{\mathrm{\infty }}}{C}_4^{k+4}{(-\frac{3x}{4})}^k$$$$=\frac{1}{1024}\underset{k=0}{\sum ^{\mathrm{\infty }}}{(-1)}^k{C}_4^{k+4}{\left(\frac{3}{4}\right)}^k{x}^k$$$$=\underset{k=0}{\sum ^{\mathrm{\infty }}}{a}_k{x}^k$$$$a_k=\frac{{(-1)}^k}{1024}{C}_4^{k+4}{\left(\frac{3}{4}\right)}^k$$$$positivecoef.whenk=even=2n$$$$negativecoef.whenk=odd=2n+1$$$$tofindthelargestcoef.weonly$$$$needtolookatpositivecoefficients,$$$$i.e.k=2n.$$$$a_{2n}=\frac{1}{1024}{C}_4^{2n+4}{\left(\frac{3}{4}\right)}^{2n}$$$$a_{2(n+1)}=\frac{1}{1024}{C}_4^{2n+6}{\left(\frac{3}{4}\right)}^{2(n+1)}$$$$tofindthelargestcoef.weneedto$$$$findthesmallestnsatisfying$$$$a_{2n}>a_{2(n+1)}$$$$i.e.$$$$\frac{1}{1024}{C}_4^{2n+4}{\left(\frac{3}{4}\right)}^{2n}>\frac{1}{1024}{C}_4^{2n+6}{\left(\frac{3}{4}\right)}^{2(n+1)}$$$$C_4^{2n+4}>C_4^{2n+6}{\left(\frac{3}{4}\right)}^2$$$$\frac{(2n+4)!}{4!\left(2n\right)!}>\frac{(2n+6)!}{4!(2n+2)!}{\left(\frac{3}{4}\right)}^2$$$$\frac{16}{9}>\frac{(n+3)(2n+5)}{(n+1)(2n+1)}$$$$16(2n^2+3n+1)>9(2n^2+11n+15)$$$$14n^2-51n-119>0$$$$n>\frac{51+\sqrt{{51}^2+4\times 14\times 119}}{28}\approx 5.3$$$$thesmallestnis6,i.e.thelargest$$$$coefficientis{a}_{12}:$$$$a_{12}=\frac{1}{1024}{C}_4^{16}{\left(\frac{3}{4}\right)}^{12}=\frac{241805655}{4294967296}$$$$$$$$youcancheckthiswithWolframAlpha:$$

Commented by mr W last updated on 13/Dec/20

Commented by aurpeyz last updated on 13/Dec/20

$$wow.thisissoexplicit.icant$$$$appreciateyouenough.thanks$$

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