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Question Number 85817 by jagoll last updated on 25/Mar/20

$$\mathrm{what}\mathrm{is}\mathrm{coefficient}\mathrm{of}{\mathrm{x}}^2\mathrm{in}$$$$\mathrm{the}\mathrm{expansion}{\left[(1-\mathrm{x})(1+2\mathrm{x})\right]}^6$$

Answered by john santu last updated on 25/Mar/20

$${(1+x-2x^2)}^6=$$$$\underset{a+b+c⩽6}{\sum ^6}\left(\begin{array}{c}6\\ a,b,c\end{array}\right)1^a{x}^b{(-2x^2)}^c$$$$\Rightarrow a+b+2c=6$$$$\left(i\right)c=1\wedge a=5$$$$\Rightarrow \frac{6!}{5!.1!}.1^5{(-2x^2)}^1=-12x^2$$$$\left(ii\right)b=2\wedge a=4$$$$\Rightarrow \frac{6!}{4!.2!}{1}^4.x^2=15x^2$$$$hencecoefficient{x}^2=-12+15=3$$

Commented by jagoll last updated on 25/Mar/20

$$\mathrm{thank}\mathrm{you}\mathrm{sir}$$

Answered by mr W last updated on 25/Mar/20

$${(1+x-2x^2)}^6$$$$coef.of{x}^2=\frac{6\times 5}{2}+6\times (-2)=3$$

Commented by jagoll last updated on 25/Mar/20

$$\mathrm{this}\mathrm{is}\mathrm{short}\mathrm{cut}\mathrm{sir}?$$

Commented by mr W last updated on 25/Mar/20

$$eachofthe6x^{\prime }swillbemultiplicated$$$$eachoftheother{x}^{\prime }stoform{x}^2,$$$$thereare{C}_2^{6}combinations,sumof$$$$thecoef.is\frac{6\times 5}{2}.$$$$eachofthe6(-2x^2)willbemultiplicated$$$$theother{1}^{\prime }s,thesumofthecoef.is$$$$6\times (-2).$$$$therefore\Rightarrow \frac{6\times 5}{2}+6\times (-2)=3.$$

Commented by jagoll last updated on 25/Mar/20

$$\mathrm{in}\mathrm{generally}\mathrm{sir}$$$$\mathrm{if}\mathrm{the}\mathrm{question}{(1+\mathrm{x}-{2\mathrm{x}}^2)}^{\mathrm{n}}$$$$\mathrm{for}\mathrm{n}\mathrm{even}?$$

Commented by jagoll last updated on 25/Mar/20

$${\mathrm{C}}_2^{\mathrm{n}}+(-2\mathrm{n})\mathrm{sir}$$

Commented by mr W last updated on 25/Mar/20

$$nisany+veinteger,{mustn}^{\prime }tbeeven.$$

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