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Question Number 134798 by bramlexs22 last updated on 07/Mar/21
$$\mathrm{Binomial}\:\mathrm{theorem} \\ $$Given that the coefficient of x^2 in the expansion of (1+ax) (3-2 x) ^5 is 1440, what is the value of the constant a?
Answered by Ñï= last updated on 07/Mar/21
$$\begin{pmatrix}{\mathrm{5}}\\{\mathrm{2}}\end{pmatrix}\left(−\mathrm{2}{x}\right)^{\mathrm{2}} \mathrm{3}^{\mathrm{3}} +{ax}\begin{pmatrix}{\mathrm{5}}\\{\mathrm{1}}\end{pmatrix}\left(−\mathrm{2}{x}\right)\mathrm{3}^{\mathrm{4}} =\mathrm{10}\left(\mathrm{108}−\mathrm{81}{a}\right){x}^{\mathrm{2}} \\ $$$$\mathrm{10}\left(\mathrm{108}−\mathrm{81}{a}\right)=\mathrm{1440} \\ $$$$\Rightarrow{a}=−\frac{\mathrm{4}}{\mathrm{9}} \\ $$
Answered by liberty last updated on 07/Mar/21
![We can find the coefficient of x^2 for f(x)= (1+ax)(3−2x)^5 by taking ((f ′′(0))/(2!)) (•) f(x)= (1+ax)(3−2x)^5 f ′(x)=a(3−2x)^5 −10(1+ax)(3−2x)^4 f ′(x)=(3−2x)^4 [ 3a−2ax−10−10ax ] f ′(x)=(3−2x)^4 (−12ax+3a−10) (••) f ′′(x)=−12a(3−2x)^4 −8(3−2x)^3 (−12ax+3a−10) f ′′(0) = −12a(81)−8(27)(3a−10) ((f ′′(0))/(2!)) = −6a(81)−4(27)(3a−10) = 1440 −486a−324a+1080=1440 −810a = 2520 ; a = −((360)/(810)) = −(4/9)](Q134802.png)
$${We}\:{can}\:{find}\:{the}\:{coefficient}\:{of}\:{x}^{\mathrm{2}} \\ $$$${for}\:{f}\left({x}\right)=\:\left(\mathrm{1}+{ax}\right)\left(\mathrm{3}−\mathrm{2}{x}\right)^{\mathrm{5}} \\ $$$${by}\:{taking}\:\frac{{f}\:''\left(\mathrm{0}\right)}{\mathrm{2}!} \\ $$$$\left(\bullet\right)\:{f}\left({x}\right)=\:\left(\mathrm{1}+{ax}\right)\left(\mathrm{3}−\mathrm{2}{x}\right)^{\mathrm{5}} \\ $$$$\:{f}\:'\left({x}\right)={a}\left(\mathrm{3}−\mathrm{2}{x}\right)^{\mathrm{5}} −\mathrm{10}\left(\mathrm{1}+{ax}\right)\left(\mathrm{3}−\mathrm{2}{x}\right)^{\mathrm{4}} \\ $$$$\:{f}\:'\left({x}\right)=\left(\mathrm{3}−\mathrm{2}{x}\right)^{\mathrm{4}} \:\left[\:\mathrm{3}{a}−\mathrm{2}{ax}−\mathrm{10}−\mathrm{10}{ax}\:\right] \\ $$$${f}\:'\left({x}\right)=\left(\mathrm{3}−\mathrm{2}{x}\right)^{\mathrm{4}} \left(−\mathrm{12}{ax}+\mathrm{3}{a}−\mathrm{10}\right) \\ $$$$ \\ $$$$\left(\bullet\bullet\right)\:{f}\:''\left({x}\right)=−\mathrm{12}{a}\left(\mathrm{3}−\mathrm{2}{x}\right)^{\mathrm{4}} −\mathrm{8}\left(\mathrm{3}−\mathrm{2}{x}\right)^{\mathrm{3}} \left(−\mathrm{12}{ax}+\mathrm{3}{a}−\mathrm{10}\right) \\ $$$$\:{f}\:''\left(\mathrm{0}\right)\:=\:−\mathrm{12}{a}\left(\mathrm{81}\right)−\mathrm{8}\left(\mathrm{27}\right)\left(\mathrm{3}{a}−\mathrm{10}\right) \\ $$$$\:\frac{{f}\:''\left(\mathrm{0}\right)}{\mathrm{2}!}\:=\:−\mathrm{6}{a}\left(\mathrm{81}\right)−\mathrm{4}\left(\mathrm{27}\right)\left(\mathrm{3}{a}−\mathrm{10}\right)\:=\:\mathrm{1440} \\ $$$$\:−\mathrm{486}{a}−\mathrm{324}{a}+\mathrm{1080}=\mathrm{1440} \\ $$$$−\mathrm{810}{a}\:=\:\mathrm{2520}\:;\:{a}\:=\:−\frac{\mathrm{360}}{\mathrm{810}}\:=\:−\frac{\mathrm{4}}{\mathrm{9}} \\ $$
Answered by mr W last updated on 07/Mar/21
$$\left(\mathrm{1}+{ax}\right)\left(\mathrm{3}−\mathrm{2}{x}\right)^{\mathrm{5}} =\left(\mathrm{1}+{ax}\right)\underset{{k}=\mathrm{0}} {\overset{\mathrm{5}} {\sum}}{C}_{{k}} ^{\mathrm{5}} \mathrm{3}^{\mathrm{5}−{k}} \left(−\mathrm{2}{x}\right)^{{k}} \\ $$$$=\left(\mathrm{1}+{ax}\right)\underset{{k}=\mathrm{0}} {\overset{\mathrm{5}} {\sum}}{C}_{{k}} ^{\mathrm{5}} \mathrm{3}^{\mathrm{5}−{k}} \left(−\mathrm{2}\right)^{{k}} {x}^{{k}} \\ $$$${coef}.\:{of}\:{x}^{\mathrm{2}} : \\ $$$${C}_{\mathrm{2}} ^{\mathrm{5}} \mathrm{3}^{\mathrm{5}−\mathrm{2}} \left(−\mathrm{2}\right)^{\mathrm{2}} +{aC}_{\mathrm{1}} ^{\mathrm{5}} \mathrm{3}^{\mathrm{5}−\mathrm{1}} \left(−\mathrm{2}\right)^{\mathrm{1}} \\ $$$$=\mathrm{1080}−\mathrm{810}{a}=\mathrm{1440} \\ $$$$\Rightarrow{a}=−\frac{\mathrm{4}}{\mathrm{9}} \\ $$