Question and Answers Forum

All Questions      Topic List

Probability and Statistics Questions

Previous in All Question      Next in All Question      

Previous in Probability and Statistics      Next in Probability and Statistics      

Question Number 80746 by jagoll last updated on 06/Feb/20

what is constan term in expansion (1+3x)^5 ((3/x)+1)^2

$${what}\:{is}\:{constan}\:{term}\:{in}\:{expansion} \\ $$$$\left(\mathrm{1}+\mathrm{3}{x}\right)^{\mathrm{5}} \left(\frac{\mathrm{3}}{{x}}+\mathrm{1}\right)^{\mathrm{2}} \\ $$

Commented by jagoll last updated on 06/Feb/20

c(x) = (((1+3x)^5 (3+x)^2 )/x^2 ) let p(x)= (1+3x)^5 (3+x)^2 p′(x)=15(3x+1)^4 (3+x)^2 +6(3+x)(1+3x)^5 constant term in c(x) = ((p′′(0))/(2!)) my method is correct?

$${c}\left({x}\right)\:=\:\frac{\left(\mathrm{1}+\mathrm{3}{x}\right)^{\mathrm{5}} \left(\mathrm{3}+{x}\right)^{\mathrm{2}} }{{x}^{\mathrm{2}} } \\ $$$${let}\:{p}\left({x}\right)=\:\left(\mathrm{1}+\mathrm{3}{x}\right)^{\mathrm{5}} \left(\mathrm{3}+{x}\right)^{\mathrm{2}} \\ $$$${p}'\left({x}\right)=\mathrm{15}\left(\mathrm{3}{x}+\mathrm{1}\right)^{\mathrm{4}} \left(\mathrm{3}+{x}\right)^{\mathrm{2}} +\mathrm{6}\left(\mathrm{3}+{x}\right)\left(\mathrm{1}+\mathrm{3}{x}\right)^{\mathrm{5}} \\ $$$${constant}\:{term}\:{in}\:{c}\left({x}\right)\:=\:\frac{{p}''\left(\mathrm{0}\right)}{\mathrm{2}!} \\ $$$${my}\:{method}\:{is}\:{correct}? \\ $$

Commented by mathmax by abdo last updated on 06/Feb/20

we have (3x+1)^5 (((x+3)/x))^2 =(1/x^2 )(3x+1)^5 (x^2 +6x +9) =(3x+1)^5 +(6/x)(3x+1)^5 +(9/x^2 )(3x+1)^5 the costant term in(3x+1)^5 is1 (6/x)(3x+1)^5 =(6/x)Σ_(k=0) ^5 C_5 ^k (3x)^k =6Σ_(k=0) ^5 3^k C_5 ^k x^(k−1) term constant ⇒k=1 ⇒c_1 =6×3^1 ×C_5 ^1 =18×5 =90 (9/x^2 )(3x+1)^5 =(9/x^2 )Σ_(k=0) ^5 C_5 ^k (3x)^k =9Σ_(k=0) ^5 3^k C_5 ^k x^(k−2) term constant ⇒k=2 ⇒c_2 =9×3^2 ×C_5 ^2 =81×((5!)/(2!3!)) =81×((5×4)/2) =810 ⇒ the constant term in (1+3x)^5 ((3/x)+1)^2 is1+90+810 =901

$${we}\:{have}\:\left(\mathrm{3}{x}+\mathrm{1}\right)^{\mathrm{5}} \left(\frac{{x}+\mathrm{3}}{{x}}\right)^{\mathrm{2}} =\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\left(\mathrm{3}{x}+\mathrm{1}\right)^{\mathrm{5}} \left({x}^{\mathrm{2}} \:+\mathrm{6}{x}\:+\mathrm{9}\right) \\ $$$$=\left(\mathrm{3}{x}+\mathrm{1}\right)^{\mathrm{5}} \:+\frac{\mathrm{6}}{{x}}\left(\mathrm{3}{x}+\mathrm{1}\right)^{\mathrm{5}} \:+\frac{\mathrm{9}}{{x}^{\mathrm{2}} }\left(\mathrm{3}{x}+\mathrm{1}\right)^{\mathrm{5}} \\ $$$${the}\:{costant}\:{term}\:{in}\left(\mathrm{3}{x}+\mathrm{1}\right)^{\mathrm{5}} {is}\mathrm{1} \\ $$$$\frac{\mathrm{6}}{{x}}\left(\mathrm{3}{x}+\mathrm{1}\right)^{\mathrm{5}} =\frac{\mathrm{6}}{{x}}\sum_{{k}=\mathrm{0}} ^{\mathrm{5}} \:{C}_{\mathrm{5}} ^{{k}} \left(\mathrm{3}{x}\right)^{{k}} \:=\mathrm{6}\sum_{{k}=\mathrm{0}} ^{\mathrm{5}} \mathrm{3}^{{k}} \:{C}_{\mathrm{5}} ^{{k}} \:{x}^{{k}−\mathrm{1}} \\ $$$${term}\:{constant}\:\Rightarrow{k}=\mathrm{1}\:\Rightarrow{c}_{\mathrm{1}} =\mathrm{6}×\mathrm{3}^{\mathrm{1}} ×{C}_{\mathrm{5}} ^{\mathrm{1}} \:=\mathrm{18}×\mathrm{5}\:=\mathrm{90} \\ $$$$\frac{\mathrm{9}}{{x}^{\mathrm{2}} }\left(\mathrm{3}{x}+\mathrm{1}\right)^{\mathrm{5}} \:=\frac{\mathrm{9}}{{x}^{\mathrm{2}} }\sum_{{k}=\mathrm{0}} ^{\mathrm{5}} \:{C}_{\mathrm{5}} ^{{k}} \left(\mathrm{3}{x}\right)^{{k}} \:=\mathrm{9}\sum_{{k}=\mathrm{0}} ^{\mathrm{5}} \:\mathrm{3}^{{k}} \:{C}_{\mathrm{5}} ^{{k}} \:{x}^{{k}−\mathrm{2}} \\ $$$${term}\:{constant}\:\Rightarrow{k}=\mathrm{2}\:\Rightarrow{c}_{\mathrm{2}} =\mathrm{9}×\mathrm{3}^{\mathrm{2}} ×{C}_{\mathrm{5}} ^{\mathrm{2}} \:=\mathrm{81}×\frac{\mathrm{5}!}{\mathrm{2}!\mathrm{3}!} \\ $$$$=\mathrm{81}×\frac{\mathrm{5}×\mathrm{4}}{\mathrm{2}}\:=\mathrm{810}\:\Rightarrow\:{the}\:{constant}\:{term}\:{in}\:\left(\mathrm{1}+\mathrm{3}{x}\right)^{\mathrm{5}} \left(\frac{\mathrm{3}}{{x}}+\mathrm{1}\right)^{\mathrm{2}} \\ $$$${is}\mathrm{1}+\mathrm{90}+\mathrm{810}\:=\mathrm{901} \\ $$

Answered by ajfour last updated on 06/Feb/20

E=(1/x^2 )(1+3x)^5 (3+x)^2 constant term in E is same as coefficient of x^2 in x^2 E = 1+[2×3×5(3)]+[9×10×9] = 1+90+810 = 901 .

$${E}=\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\left(\mathrm{1}+\mathrm{3}{x}\right)^{\mathrm{5}} \left(\mathrm{3}+{x}\right)^{\mathrm{2}} \\ $$$$\:\:{constant}\:{term}\:{in}\:{E}\:{is}\:{same}\:{as} \\ $$$${coefficient}\:{of}\:{x}^{\mathrm{2}} \:{in}\:{x}^{\mathrm{2}} {E} \\ $$$$\:\:\:=\:\mathrm{1}+\left[\mathrm{2}×\mathrm{3}×\mathrm{5}\left(\mathrm{3}\right)\right]+\left[\mathrm{9}×\mathrm{10}×\mathrm{9}\right] \\ $$$$\:\:\:=\:\mathrm{1}+\mathrm{90}+\mathrm{810}\:=\:\mathrm{901}\:. \\ $$

Commented by john santu last updated on 06/Feb/20

p(x)= (((243x^5 +...+90x^2 +15x+1)(x^2 +6x+9))/x^2 ) the constant in p(x) is ((90x^2 .9+15x.6x+1.x^2 )/x^2 )= 810+90+1=901

$${p}\left({x}\right)=\:\frac{\left(\mathrm{243}{x}^{\mathrm{5}} +...+\mathrm{90}{x}^{\mathrm{2}} +\mathrm{15}{x}+\mathrm{1}\right)\left({x}^{\mathrm{2}} +\mathrm{6}{x}+\mathrm{9}\right)}{{x}^{\mathrm{2}} } \\ $$$${the}\:{constant}\:{in}\:{p}\left({x}\right)\:{is}\: \\ $$$$\frac{\mathrm{90}{x}^{\mathrm{2}} .\mathrm{9}+\mathrm{15}{x}.\mathrm{6}{x}+\mathrm{1}.{x}^{\mathrm{2}} }{{x}^{\mathrm{2}} }= \\ $$$$\mathrm{810}+\mathrm{90}+\mathrm{1}=\mathrm{901} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com