Question and Answers Forum

All Questions      Topic List

Differential Equation Questions

Previous in All Question      Next in All Question      

Previous in Differential Equation      Next in Differential Equation      

Question Number 119464 by Engr_Jidda last updated on 24/Oct/20

Expand as far as the term in x^3 (1) (3−2x−x^2 )^9 (2) (1−x+x^2 )^8

$$\mathrm{Expand}\:\mathrm{as}\:\mathrm{far}\:\mathrm{as}\:\mathrm{the}\:\mathrm{term}\:\mathrm{in}\:\mathrm{x}^{\mathrm{3}} \\ $$$$\left(\mathrm{1}\right)\:\left(\mathrm{3}−\mathrm{2x}−\mathrm{x}^{\mathrm{2}} \right)^{\mathrm{9}} \\ $$$$\left(\mathrm{2}\right)\:\left(\mathrm{1}−\mathrm{x}+\mathrm{x}^{\mathrm{2}} \right)^{\mathrm{8}} \\ $$

Answered by TANMAY PANACEA last updated on 24/Oct/20

1)(3−2x−x^2 )^9 =a_0 +a_1 x+a_2 x^2 +a_3 x^3 +...+Px^(18) put x=0 so a_0 =3^9 diffdrntiate and put x=0 9(3−2x−x^2 )^8 (−2−2x)=0+a_1 +2a_2 x+3a_3 x^2 +..+18px^(17) put x=0 9×3^8 ×(−2)=a_1 9(3−2x−x^2 )^8 (−2)+9×8×(3−2x−x^2 )(−2−2x) =0+0+2a_2 +terms containing x put x=0 2a_2 =9×3^8 ×(−2)+9×8×(3)(−2) a_2 =−9×3^8 −9×8×3 9×(−2)×8(3−2x−x^2 )^7 ×(−2−2x)+72(−2−2x)(−2−2x)+72(3−2x−x^2 )(−2) =0+0+0+6a_3 +terms containing x put x=0 6a_3 =9×(−2)×8×3^7 ×(−2)+72×(−2)(−2)+72×3×−2 a_3 =9×(−1)×8×3^6 ×(−2)+12×4+72×−1 Thus wd get value a_0 , a_1 , a_2 , and a_3

$$\left.\mathrm{1}\right)\left(\mathrm{3}−\mathrm{2}{x}−{x}^{\mathrm{2}} \right)^{\mathrm{9}} \\ $$$$={a}_{\mathrm{0}} +{a}_{\mathrm{1}} {x}+{a}_{\mathrm{2}} {x}^{\mathrm{2}} +{a}_{\mathrm{3}} {x}^{\mathrm{3}} +...+{Px}^{\mathrm{18}} \\ $$$${put}\:{x}=\mathrm{0}\:{so}\:{a}_{\mathrm{0}} =\mathrm{3}^{\mathrm{9}} \\ $$$${diffdrntiate}\:{and}\:{put}\:{x}=\mathrm{0} \\ $$$$\mathrm{9}\left(\mathrm{3}−\mathrm{2}{x}−{x}^{\mathrm{2}} \right)^{\mathrm{8}} \left(−\mathrm{2}−\mathrm{2}{x}\right)=\mathrm{0}+{a}_{\mathrm{1}} +\mathrm{2}{a}_{\mathrm{2}} {x}+\mathrm{3}{a}_{\mathrm{3}} {x}^{\mathrm{2}} +..+\mathrm{18}{px}^{\mathrm{17}} \\ $$$${put}\:{x}=\mathrm{0} \\ $$$$\mathrm{9}×\mathrm{3}^{\mathrm{8}} ×\left(−\mathrm{2}\right)={a}_{\mathrm{1}} \\ $$$$\mathrm{9}\left(\mathrm{3}−\mathrm{2}{x}−{x}^{\mathrm{2}} \right)^{\mathrm{8}} \left(−\mathrm{2}\right)+\mathrm{9}×\mathrm{8}×\left(\mathrm{3}−\mathrm{2}{x}−{x}^{\mathrm{2}} \right)\left(−\mathrm{2}−\mathrm{2}{x}\right) \\ $$$$=\mathrm{0}+\mathrm{0}+\mathrm{2}{a}_{\mathrm{2}} +{terms}\:{containing}\:{x} \\ $$$${put}\:{x}=\mathrm{0} \\ $$$$\mathrm{2}{a}_{\mathrm{2}} =\mathrm{9}×\mathrm{3}^{\mathrm{8}} ×\left(−\mathrm{2}\right)+\mathrm{9}×\mathrm{8}×\left(\mathrm{3}\right)\left(−\mathrm{2}\right) \\ $$$${a}_{\mathrm{2}} =−\mathrm{9}×\mathrm{3}^{\mathrm{8}} −\mathrm{9}×\mathrm{8}×\mathrm{3} \\ $$$$\mathrm{9}×\left(−\mathrm{2}\right)×\mathrm{8}\left(\mathrm{3}−\mathrm{2}{x}−{x}^{\mathrm{2}} \right)^{\mathrm{7}} ×\left(−\mathrm{2}−\mathrm{2}{x}\right)+\mathrm{72}\left(−\mathrm{2}−\mathrm{2}{x}\right)\left(−\mathrm{2}−\mathrm{2}{x}\right)+\mathrm{72}\left(\mathrm{3}−\mathrm{2}{x}−{x}^{\mathrm{2}} \right)\left(−\mathrm{2}\right) \\ $$$$=\mathrm{0}+\mathrm{0}+\mathrm{0}+\mathrm{6}{a}_{\mathrm{3}} +{terms}\:{containing}\:{x} \\ $$$${put}\:{x}=\mathrm{0} \\ $$$$\mathrm{6}{a}_{\mathrm{3}} =\mathrm{9}×\left(−\mathrm{2}\right)×\mathrm{8}×\mathrm{3}^{\mathrm{7}} ×\left(−\mathrm{2}\right)+\mathrm{72}×\left(−\mathrm{2}\right)\left(−\mathrm{2}\right)+\mathrm{72}×\mathrm{3}×−\mathrm{2} \\ $$$${a}_{\mathrm{3}} =\mathrm{9}×\left(−\mathrm{1}\right)×\mathrm{8}×\mathrm{3}^{\mathrm{6}} ×\left(−\mathrm{2}\right)+\mathrm{12}×\mathrm{4}+\mathrm{72}×−\mathrm{1} \\ $$$$\boldsymbol{{T}}{hus}\:{wd}\:{get}\:{value}\:{a}_{\mathrm{0}} \:,\:{a}_{\mathrm{1}} ,\:\:{a}_{\mathrm{2}} ,\:{and}\:{a}_{\mathrm{3}} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$

Commented by TANMAY PANACEA last updated on 24/Oct/20

I have lethergy[to calculate the vslue of a_0 ,a_1 ..a_3 ...

$$\boldsymbol{{I}}\:{have}\:{lethergy}\left[{to}\:{calculate}\:{the}\:{vslue}\:{of}\:{a}_{\mathrm{0}} ,{a}_{\mathrm{1}} ..{a}_{\mathrm{3}} ...\right. \\ $$

Commented by Engr_Jidda last updated on 24/Oct/20

thank you so much sir

$$\mathrm{thank}\:\mathrm{you}\:\mathrm{so}\:\mathrm{much}\:\mathrm{sir} \\ $$

Answered by TANMAY PANACEA last updated on 24/Oct/20

2) (1−x+x^2 )^9 =a_0 +a_1 x+a_2 x^2 +a_3 x^3 +...+Qx^(18) step i)put x=0 a_0 =1 ii)differentiate both side and put x=0 get value of a_1 ... ...Thus we get a_(0 ) ,a_1 ,a_2 and a_3 ...

$$\left.\mathrm{2}\right) \\ $$$$\left(\mathrm{1}−{x}+{x}^{\mathrm{2}} \right)^{\mathrm{9}} ={a}_{\mathrm{0}} +{a}_{\mathrm{1}} {x}+{a}_{\mathrm{2}} {x}^{\mathrm{2}} +{a}_{\mathrm{3}} {x}^{\mathrm{3}} +...+{Qx}^{\mathrm{18}} \\ $$$${step}\: \\ $$$$\left.{i}\right){put}\:{x}=\mathrm{0}\:\:\:{a}_{\mathrm{0}} =\mathrm{1} \\ $$$$\left.{ii}\right){differentiate}\:{both}\:{side}\:{and}\:{put}\:{x}=\mathrm{0}\:{get} \\ $$$${value}\:{of}\:{a}_{\mathrm{1}} \\ $$$$... \\ $$$$...{Thus}\:{we}\:{get}\:{a}_{\mathrm{0}\:} ,{a}_{\mathrm{1}} ,{a}_{\mathrm{2}} \:{and}\:{a}_{\mathrm{3}} ... \\ $$

Commented by Engr_Jidda last updated on 24/Oct/20

thank you sir

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

Commented by TANMAY PANACEA last updated on 24/Oct/20

most welcome sir

$${most}\:{welcome}\:{sir} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com