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The-pisitive-value-of-a-so-that-the-coefficient-of-x-5-and-x-15-are-equal-in-the-expansion-of-x-2-a-x-3-10-

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Question Number 56138 by gunawan last updated on 11/Mar/19
The pisitive value of a so that the coefficient of x^5 and x^(15) are equal in the expansion of (x^2 + (a/x^3 ))^(10)
$$\mathrm{The}\:\mathrm{pisitive}\:\mathrm{value}\:\mathrm{of}\:\:{a}\:\mathrm{so}\:\mathrm{that}\:\mathrm{the} \\ $$$$\mathrm{coefficient}\:\mathrm{of}\:{x}^{\mathrm{5}} \:\mathrm{and}\:{x}^{\mathrm{15}} \:\mathrm{are}\:\mathrm{equal}\:\mathrm{in}\:\mathrm{the} \\ $$$$\mathrm{expansion}\:\mathrm{of}\:\left({x}^{\mathrm{2}} +\:\frac{{a}}{{x}^{\mathrm{3}} }\right)^{\mathrm{10}} \\ $$
Commented by maxmathsup by imad last updated on 11/Mar/19
we have (x^2 +(a/x^3 ))^(10) =Σ_(k=0) ^(10) C_(10) ^k (x^2 )^k (ax^(−3) )^(10−k) =Σ_(k=0) ^(10) C_(10) ^k x^(2k) a^(10−k) x^(−30+3k) =Σ_(k=0) ^(10) C_(10) ^k a^(10−k) x^(5k−30) the coefficient of x^5 is λ_5 /5k−30=5 ⇒k =7 ⇒ λ_5 = C_(10) ^7 a^3 also the coefgicient of x^(15) is λ_(15) /5k−30=15 ⇒5k=45 ⇒k=9 ⇒λ_(15) =C_(10) ^9 a so λ_5 =λ_(15) ⇒C_(10) ^7 a^3 =C_(10) ^9 a ⇒ ((10!)/(7!3!)) a^3 =10a ⇒((10 .9.8)/6)a^3 =10 a ⇒((3.3.4.2)/(3.2)) a^3 =a ⇒12a^3 −a =0 ⇒ a(12a^2 −1) =0 ⇒a=0 or 12a^2 =1 ⇒a=0 or a =+^− (1/(2(√3))) .
$${we}\:{have}\:\left({x}^{\mathrm{2}} \:+\frac{{a}}{{x}^{\mathrm{3}} }\right)^{\mathrm{10}} \:=\sum_{{k}=\mathrm{0}} ^{\mathrm{10}} \:{C}_{\mathrm{10}} ^{{k}} \:\left({x}^{\mathrm{2}} \right)^{{k}} \left({ax}^{−\mathrm{3}} \right)^{\mathrm{10}−{k}} \\ $$$$=\sum_{{k}=\mathrm{0}} ^{\mathrm{10}} \:{C}_{\mathrm{10}} ^{{k}} \:{x}^{\mathrm{2}{k}} \:{a}^{\mathrm{10}−{k}} \:{x}^{−\mathrm{30}+\mathrm{3}{k}} \:=\sum_{{k}=\mathrm{0}} ^{\mathrm{10}} \:{C}_{\mathrm{10}} ^{{k}} \:{a}^{\mathrm{10}−{k}} \:{x}^{\mathrm{5}{k}−\mathrm{30}} \\ $$$${the}\:{coefficient}\:{of}\:{x}^{\mathrm{5}} \:{is}\:\lambda_{\mathrm{5}} \:/\mathrm{5}{k}−\mathrm{30}=\mathrm{5}\:\Rightarrow{k}\:=\mathrm{7}\:\Rightarrow \\ $$$$\lambda_{\mathrm{5}} =\:{C}_{\mathrm{10}} ^{\mathrm{7}} \:{a}^{\mathrm{3}} \:\:\:\:{also}\:{the}\:{coefgicient}\:{of}\:{x}^{\mathrm{15}} \:{is}\:\lambda_{\mathrm{15}} /\mathrm{5}{k}−\mathrm{30}=\mathrm{15}\:\Rightarrow\mathrm{5}{k}=\mathrm{45}\:\Rightarrow{k}=\mathrm{9} \\ $$$$\Rightarrow\lambda_{\mathrm{15}} ={C}_{\mathrm{10}} ^{\mathrm{9}} \:{a}\:\:{so}\:\lambda_{\mathrm{5}} =\lambda_{\mathrm{15}} \:\Rightarrow{C}_{\mathrm{10}} ^{\mathrm{7}} {a}^{\mathrm{3}} ={C}_{\mathrm{10}} ^{\mathrm{9}} \:{a}\:\Rightarrow \\ $$$$\frac{\mathrm{10}!}{\mathrm{7}!\mathrm{3}!}\:{a}^{\mathrm{3}} \:=\mathrm{10}{a}\:\Rightarrow\frac{\mathrm{10}\:.\mathrm{9}.\mathrm{8}}{\mathrm{6}}{a}^{\mathrm{3}} \:=\mathrm{10}\:{a}\:\Rightarrow\frac{\mathrm{3}.\mathrm{3}.\mathrm{4}.\mathrm{2}}{\mathrm{3}.\mathrm{2}}\:{a}^{\mathrm{3}} \:={a}\:\Rightarrow\mathrm{12}{a}^{\mathrm{3}} −{a}\:=\mathrm{0}\:\Rightarrow \\ $$$${a}\left(\mathrm{12}{a}^{\mathrm{2}} −\mathrm{1}\right)\:=\mathrm{0}\:\Rightarrow{a}=\mathrm{0}\:{or}\:\mathrm{12}{a}^{\mathrm{2}} =\mathrm{1}\:\Rightarrow{a}=\mathrm{0}\:{or}\:{a}\:=\overset{−} {+}\:\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{3}}}\:. \\ $$
Commented by maxmathsup by imad last updated on 11/Mar/19
but a>0 ⇒a =(1/(2(√3))) .
$${but}\:{a}>\mathrm{0}\:\Rightarrow{a}\:=\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{3}}}\:. \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 11/Mar/19
r+1 th term contains x^5 10c_r (x^2 )^(10−r) ((a/x^3 ))^r =10c_r ×x^(20−2r−3r) ×a^r given 20−5r=5 5r=15→r=3 10c_3 ×x^(20−5×3) ×a^3 →((10!)/(3!7!))×a^3 ×x^5 ... r_1 +1th term contsins x^(15) 10c_r_1 ×x^(20−5r_1 ) ×a^r_1 20−5r_1 =15→5r_1 =5 r_1 =1 10c_1 ×x^(15) ×a^1 →((10!)/(1!9!))×a^1 ×x^(15) according to question ((10!)/(3!7!))a^3 =((10!)/(1!9!))a^1 a^2 =((3×2×7!)/(9×8×7!))=(1/(3×4)) a=(1/(2(√3)))
$${r}+\mathrm{1}\:{th}\:{term}\:\:{contains}\:{x}^{\mathrm{5}} \\ $$$$\mathrm{10}{c}_{{r}} \left({x}^{\mathrm{2}} \right)^{\mathrm{10}−{r}} \left(\frac{{a}}{{x}^{\mathrm{3}} }\right)^{{r}} \\ $$$$=\mathrm{10}{c}_{{r}} ×{x}^{\mathrm{20}−\mathrm{2}{r}−\mathrm{3}{r}} ×{a}^{{r}} \\ $$$${given}\:\mathrm{20}−\mathrm{5}{r}=\mathrm{5} \\ $$$$\mathrm{5}{r}=\mathrm{15}\rightarrow{r}=\mathrm{3} \\ $$$$\mathrm{10}{c}_{\mathrm{3}} ×{x}^{\mathrm{20}−\mathrm{5}×\mathrm{3}} ×{a}^{\mathrm{3}} \rightarrow\frac{\mathrm{10}!}{\mathrm{3}!\mathrm{7}!}×{a}^{\mathrm{3}} ×{x}^{\mathrm{5}} \\ $$$$… \\ $$$${r}_{\mathrm{1}} +\mathrm{1}{th}\:{term}\:{contsins}\:{x}^{\mathrm{15}} \\ $$$$\mathrm{10}{c}_{{r}_{\mathrm{1}} } ×{x}^{\mathrm{20}−\mathrm{5}{r}_{\mathrm{1}} } ×{a}^{{r}_{\mathrm{1}} } \\ $$$$\mathrm{20}−\mathrm{5}{r}_{\mathrm{1}} =\mathrm{15}\rightarrow\mathrm{5}{r}_{\mathrm{1}} =\mathrm{5}\:\:\:\:{r}_{\mathrm{1}} =\mathrm{1} \\ $$$$\mathrm{10}{c}_{\mathrm{1}} ×{x}^{\mathrm{15}} ×{a}^{\mathrm{1}} \rightarrow\frac{\mathrm{10}!}{\mathrm{1}!\mathrm{9}!}×{a}^{\mathrm{1}} ×{x}^{\mathrm{15}} \\ $$$${according}\:{to}\:{question} \\ $$$$\frac{\mathrm{10}!}{\mathrm{3}!\mathrm{7}!}{a}^{\mathrm{3}} =\frac{\mathrm{10}!}{\mathrm{1}!\mathrm{9}!}{a}^{\mathrm{1}} \\ $$$${a}^{\mathrm{2}} =\frac{\mathrm{3}×\mathrm{2}×\mathrm{7}!}{\mathrm{9}×\mathrm{8}×\mathrm{7}!}=\frac{\mathrm{1}}{\mathrm{3}×\mathrm{4}} \\ $$$${a}=\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{3}}} \\ $$

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