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

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