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Question Number 134798 by bramlexs22 last updated on 07/Mar/21

Binomial theorem

$$\mathrm{Binomial}\:\mathrm{theorem} \\ $$ 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?\\n

Answered by Ñï= last updated on 07/Mar/21

((5),(2) )(−2x)^2 3^3 +ax ((5),(1) )(−2x)3^4 =10(108−81a)x^2 10(108−81a)=1440 ⇒a=−(4/9)

$$\begin{pmatrix}{\mathrm{5}}\\{\mathrm{2}}\end{pmatrix}\left(−\mathrm{2}{x}\right)^{\mathrm{2}} \mathrm{3}^{\mathrm{3}} +{ax}\begin{pmatrix}{\mathrm{5}}\\{\mathrm{1}}\end{pmatrix}\left(−\mathrm{2}{x}\right)\mathrm{3}^{\mathrm{4}} =\mathrm{10}\left(\mathrm{108}−\mathrm{81}{a}\right){x}^{\mathrm{2}} \\ $$ $$\mathrm{10}\left(\mathrm{108}−\mathrm{81}{a}\right)=\mathrm{1440} \\ $$ $$\Rightarrow{a}=−\frac{\mathrm{4}}{\mathrm{9}} \\ $$

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)

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Answered by mr W last updated on 07/Mar/21

(1+ax)(3−2x)^5 =(1+ax)Σ_(k=0) ^5 C_k ^5 3^(5−k) (−2x)^k =(1+ax)Σ_(k=0) ^5 C_k ^5 3^(5−k) (−2)^k x^k coef. of x^2 : C_2 ^5 3^(5−2) (−2)^2 +aC_1 ^5 3^(5−1) (−2)^1 =1080−810a=1440 ⇒a=−(4/9)

$$\left(\mathrm{1}+{ax}\right)\left(\mathrm{3}−\mathrm{2}{x}\right)^{\mathrm{5}} =\left(\mathrm{1}+{ax}\right)\underset{{k}=\mathrm{0}} {\overset{\mathrm{5}} {\sum}}{C}_{{k}} ^{\mathrm{5}} \mathrm{3}^{\mathrm{5}−{k}} \left(−\mathrm{2}{x}\right)^{{k}} \\ $$ $$=\left(\mathrm{1}+{ax}\right)\underset{{k}=\mathrm{0}} {\overset{\mathrm{5}} {\sum}}{C}_{{k}} ^{\mathrm{5}} \mathrm{3}^{\mathrm{5}−{k}} \left(−\mathrm{2}\right)^{{k}} {x}^{{k}} \\ $$ $${coef}.\:{of}\:{x}^{\mathrm{2}} : \\ $$ $${C}_{\mathrm{2}} ^{\mathrm{5}} \mathrm{3}^{\mathrm{5}−\mathrm{2}} \left(−\mathrm{2}\right)^{\mathrm{2}} +{aC}_{\mathrm{1}} ^{\mathrm{5}} \mathrm{3}^{\mathrm{5}−\mathrm{1}} \left(−\mathrm{2}\right)^{\mathrm{1}} \\ $$ $$=\mathrm{1080}−\mathrm{810}{a}=\mathrm{1440} \\ $$ $$\Rightarrow{a}=−\frac{\mathrm{4}}{\mathrm{9}} \\ $$

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