Question Number 56137 by gunawan last updated on 11/Mar/19
$$\mathrm{If}\:\left(\mathrm{1}+\mathrm{2}{x}+{x}^{\mathrm{2}} \right)^{{n}} \:=\:\underset{{r}=\mathrm{0}} {\overset{\mathrm{2}{n}} {\sum}}\:{a}_{{r}} \:{x}^{{r}} ,\:\mathrm{then}\:{a}_{{r}} = \\ $$
Commented by maxmathsup by imad last updated on 11/Mar/19
![(1+2x+x^2 )^n =(x+1)^(2n) =Σ_(r=0) ^(2n) C_(2n) ^r x^r ⇒a_r =C_(2n) ^r =(((2n)!)/(r!(2n−r)!)) and r∈[[0,2n]].](https://www.tinkutara.com/question/Q56181.png)
$$\left(\mathrm{1}+\mathrm{2}{x}+{x}^{\mathrm{2}} \right)^{{n}} =\left({x}+\mathrm{1}\right)^{\mathrm{2}{n}} \:=\sum_{{r}=\mathrm{0}} ^{\mathrm{2}{n}} \:{C}_{\mathrm{2}{n}} ^{{r}} \:{x}^{{r}} \:\Rightarrow{a}_{{r}} ={C}_{\mathrm{2}{n}} ^{{r}} \:=\frac{\left(\mathrm{2}{n}\right)!}{{r}!\left(\mathrm{2}{n}−{r}\right)!} \\ $$$${and}\:{r}\in\left[\left[\mathrm{0},\mathrm{2}{n}\right]\right]. \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 11/Mar/19
$$\left(\mathrm{1}+\mathrm{2}{x}+{x}^{\mathrm{2}} \right)^{{n}} \\ $$$$\left(\mathrm{1}+{x}\right)^{\mathrm{2}{n}} ={a}_{\mathrm{0}} {x}^{\mathrm{0}} +{a}_{\mathrm{1}} {x}^{\mathrm{1}} +{a}_{\mathrm{2}} {x}^{\mathrm{2}} +…+{a}_{\mathrm{2}{n}} {x}^{\mathrm{2}{n}} \\ $$$${put}\:\:{x}=\mathrm{1} \\ $$$$\mathrm{2}^{\mathrm{2}{n}} =\underset{{r}=\mathrm{0}} {\overset{\mathrm{2}{n}} {\sum}}{a}_{{r}} \\ $$$${r}+\mathrm{1}\:{the}\:{term}\:\:\:\mathrm{2}{nc}_{{r}} {x}^{\mathrm{2}{n}} \\ $$$${so}\:\:\:{a}_{{r}} =\mathrm{2}{nc}_{{r}} =\frac{\mathrm{2}{n}!}{{r}!\left(\mathrm{2}{n}−{r}\right)!} \\ $$