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Question Number 114739 by mr W last updated on 20/Sep/20

$$findthelargestandsmallest$$$$coefficientin{(4+3x)}^{-5}.$$

Answered by mr W last updated on 20/Sep/20

$${(4+3x)}^{-5}=4^{-5}{(1+\frac{3x}{4})}^{-5}$$$$=4^{-5}\underset{k=0}{\sum ^{\mathrm{\infty }}}{(-1)}^k{C}_4^{k+4}{\left(\frac{3}{4}\right)}^k{x}^k$$$$weseetheeventermshavepositive$$$$coefficientsandtheoddtermshave$$$$alwaysnegativecoefficients.$$$$$$$$eventerms:k=2nwithn=0,1,2,3,...$$$$a_{2n}=4^{-5}{C}_4^{2n+4}{\left(\frac{3}{4}\right)}^{2n}$$$$let{a}_{2n}>a_{2(n+1)}$$$$4^{-5}{C}_4^{2n+4}{\left(\frac{3}{4}\right)}^{2n}>4^{-5}{C}_4^{2(n+1)+4}{\left(\frac{3}{4}\right)}^{2(n+1)}$$$$\frac{(2n+4)!}{4!\left(2n\right)!}>\frac{(2n+6)!}{4!(2n+2)!}{\left(\frac{3}{4}\right)}^2$$$$\frac{16}{9}>\frac{(2n+5)(n+3)}{(2n+1)(n+1)}$$$$15n^2-51n-119>0$$$$n>\frac{51+\sqrt{{51}^2+4\times 15\times 119}}{30}\approx 4.99$$$$\Rightarrow n⩾5,i.e.thelargestcoefficient$$$$isfortheterm{x}^{10}:$$$$a_{10}=4^{-5}{C}_4^{14}{\left(\frac{3}{4}\right)}^{10}=\frac{1001\times 3^{10}}{4^{15}}$$$$=\frac{59108049}{1073741824}$$$$$$$$oddterms:k=2n+1$$$$a_{2n+1}=-4^{-5}{C}_4^{2n+5}{\left(\frac{3}{4}\right)}^{2n+1}$$$$let{a}_{2n+1}<a_{2(n+1)+1}$$$$-4^{-5}{C}_4^{2n+5}{\left(\frac{3}{4}\right)}^{2n+1}<-4^{-5}{C}_4^{2n+7}{\left(\frac{3}{4}\right)}^{2n+3}$$$$C_4^{2n+5}>C_4^{2n+7}{\left(\frac{3}{4}\right)}^2$$$$\frac{16}{9}>\frac{(n+3)(2n+7)}{(n+1)(2n+3)}$$$$14n^2-37n-141>0$$$$n>\frac{37+\sqrt{{37}^2+4\times 14\times 141}}{28}\approx 4.8$$$$\Rightarrow n⩾5,i.e.thesmallestcoefficient$$$$isfortheterm{x}^{11}.$$$$a_{11}=-4^{-5}{C}_4^{15}{\left(\frac{3}{4}\right)}^{11}=-\frac{1365\times 3^{11}}{4^{16}}$$$$=-\frac{241805655}{4294967396}$$

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