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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-




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(10881a)x210(10881a)=1440a=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)(32x)5bytakingf(0)2!()f(x)=(1+ax)(32x)5f(x)=a(32x)510(1+ax)(32x)4f(x)=(32x)4[3a2ax1010ax]f(x)=(32x)4(12ax+3a10)()f(x)=12a(32x)48(32x)3(12ax+3a10)f(0)=12a(81)8(27)(3a10)f(0)2!=6a(81)4(27)(3a10)=1440486a324a+1080=1440810a=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)(32x)5=(1+ax)5k=0Ck535k(2x)k=(1+ax)5k=0Ck535k(2)kxkcoef.ofx2:C25352(2)2+aC15351(2)1=1080810a=1440a=49

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