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Question Number 42684 by 1996ajaykmar@email.com last updated on 31/Aug/18

$$\mathrm{The}\mathrm{coefficient}\mathrm{of}{x}^4\mathrm{in}\mathrm{the}\mathrm{expansion}\mathrm{of}$$$${(\frac{x}{2}-\frac{3}{x^2})}^{10}\mathrm{is}$$

Commented by maxmathsup by imad last updated on 31/Aug/18

$$letp\left(x\right)={(\frac{x}{2}-\frac{3}{x^2})}^{10}\Rightarrow p\left(x\right)=\frac{1}{2^{10}{x}^{20}}{(x^3-6)}^{10}$$$$=2^{-10}{x}^{-20}\sum _{k=0}^{10}{C}_{10}^k{x}^{3k}{(-6)}^{10-k}=2^{-10}\sum _{k=0}^{10}{C}_{10}^k{x}^{3k-20}{(-6)}^{10-3k}$$$$thecoefficient{x}^{4}isobtainedwhen3k-20=4\Rightarrow k=8\Rightarrow$$$${\lambda }_4=2^{-10}{C}_{10}^8{(-6)}^{10-24}=2^{-10}{(-6)}^{-14}{C}_{10}^8=\frac{2^{-10}}{6^{14}}{C}_{10}^8$$$$=\frac{2^{-10}}{2^{14}.3^{14}}\frac{10!}{8!2!}=\frac{1}{2^{24}.3^{14}}\left(45\right)=\frac{45}{2^{24}{3}^{13}}=\frac{3^2.5}{2^{24}{3}^{13}}=\frac{5}{2^{24}.3^{11}}.$$

Commented by Rio Michael last updated on 31/Aug/18

$${(\frac{x}{2}-\frac{3}{x^2})}^{10}$$$$Generalterm={}^n{C}_r.X^{n-r}.a^r$$$$for{(\frac{x}{2}-\frac{3}{x^2})}^{10}={}^{10}{C}_r.{\left(\frac{x}{2}\right)}^{10-r}{(-\frac{3}{x^2})}^r$$$$={}^{10}{C}_r.x^{10-r}.2^{-10+r}.-3^r.x^{-r}$$$$={}^{10}{C}_r.2^{-10+r}.-3^r.x^{10-r}.x^{-r}$$$${=}^{10}{C}_r.2^{-10+r}.-3^r.x^{10-2r}$$$$fortermin{x}^4,x^4=x^{10-2r}$$$$\Rightarrow 4=10-2r$$$$-6=-2r$$$$r=3$$$$termin{x}^4={}^{10}{C}_3.2^{-7}-3^3$$$$=\left(120\right)\left(\frac{1}{128}\right)(-27)$$$$=-\frac{405}{16}$$

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