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

Binomial 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?

BinomialtheoremGiven 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

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

(52)(−2x)233+ax(51)(−2x)34=10(108−81a)x210(108−81a)=1440⇒a=−49

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)

Wecanfindthecoefficientofx2forf(x)=(1+ax)(3−2x)5bytakingf″(0)2!(∙)f(x)=(1+ax)(3−2x)5f′(x)=a(3−2x)5−10(1+ax)(3−2x)4f′(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=−360810=−49

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)

(1+ax)(3−2x)5=(1+ax)∑5k=0Ck535−k(−2x)k=(1+ax)∑5k=0Ck535−k(−2)kxkcoef.ofx2:C2535−2(−2)2+aC1535−1(−2)1=1080−810a=1440⇒a=−49

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