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Question Number 54876 by shaddie last updated on 13/Feb/19

find the coefficientof x^2 in the binomial expansion of (x^2 +(2/x))^4

$$\mathrm{find}\:\mathrm{the}\:\mathrm{coefficientof}\:\mathrm{x}^{\mathrm{2}} \:\mathrm{in}\:\mathrm{the}\:\mathrm{binomial} \\ $$$$\mathrm{expansion}\:\mathrm{of}\:\left(\mathrm{x}^{\mathrm{2}} +\frac{\mathrm{2}}{\mathrm{x}}\right)^{\mathrm{4}} \\ $$

Commented by maxmathsup by imad last updated on 13/Feb/19

we have (x^2 +(2/x))^4 =Σ_(k=0) ^4 C_4 ^k (x^2 )^k ((2/x))^(4−k) =Σ_(k=0) ^4 C_4 ^k x^(2k) 2^(4−k) x^(k−4) =Σ_(k=0) ^4 C_4 ^k 2^(4−k) x^(3k−4) the coefficient of x^2 is λ_k / 3k−4 =2 ⇒3k =6 ⇒k =2 ⇒λ_k =λ_2 = C_4 ^2 2^(4−2) =4 C_4 ^2 C_4 ^2 =((4!)/(2!2!)) =((4×3 )/2) =6 ⇒λ_2 =4×6 =24 .

$${we}\:{have}\:\left({x}^{\mathrm{2}} \:+\frac{\mathrm{2}}{{x}}\right)^{\mathrm{4}} \:=\sum_{{k}=\mathrm{0}} ^{\mathrm{4}} \:{C}_{\mathrm{4}} ^{{k}} \:\:\:\left({x}^{\mathrm{2}} \right)^{{k}} \left(\frac{\mathrm{2}}{{x}}\right)^{\mathrm{4}−{k}} \\ $$$$=\sum_{{k}=\mathrm{0}} ^{\mathrm{4}} \:{C}_{\mathrm{4}} ^{{k}} \:\:{x}^{\mathrm{2}{k}} \:\:\mathrm{2}^{\mathrm{4}−{k}} \:{x}^{{k}−\mathrm{4}} \:=\sum_{{k}=\mathrm{0}} ^{\mathrm{4}} \:{C}_{\mathrm{4}} ^{{k}} \:\:\mathrm{2}^{\mathrm{4}−{k}} \:{x}^{\mathrm{3}{k}−\mathrm{4}} \:\:\:\:{the}\:{coefficient}\:{of}\:{x}^{\mathrm{2}} \:{is} \\ $$$$\lambda_{{k}} \:/\:\:\mathrm{3}{k}−\mathrm{4}\:=\mathrm{2}\:\Rightarrow\mathrm{3}{k}\:=\mathrm{6}\:\Rightarrow{k}\:=\mathrm{2}\:\Rightarrow\lambda_{{k}} =\lambda_{\mathrm{2}} =\:{C}_{\mathrm{4}} ^{\mathrm{2}} \:\mathrm{2}^{\mathrm{4}−\mathrm{2}} \:\:=\mathrm{4}\:{C}_{\mathrm{4}} ^{\mathrm{2}} \\ $$$${C}_{\mathrm{4}} ^{\mathrm{2}} \:=\frac{\mathrm{4}!}{\mathrm{2}!\mathrm{2}!}\:=\frac{\mathrm{4}×\mathrm{3}\:}{\mathrm{2}}\:=\mathrm{6}\:\Rightarrow\lambda_{\mathrm{2}} =\mathrm{4}×\mathrm{6}\:=\mathrm{24}\:. \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 14/Feb/19

let r+1 th term contains x^2 4c_r (x^2 )^(4−r) ((2/x))^r 4c_r ×x^(8−2r) ×(2^r /x^r ) hence x^(8−2r−r) =x^2 −3r=2−8 →−3r=−6 r=2 4c_2 ×x^(8−3×2) ×2^2 → ((4!)/(2!2!))×4×x^2 so the coefficient is →6×4=24

$${let}\:{r}+\mathrm{1}\:{th}\:{term}\:{contains}\:{x}^{\mathrm{2}} \\ $$$$\mathrm{4}{c}_{{r}} \left({x}^{\mathrm{2}} \right)^{\mathrm{4}−{r}} \left(\frac{\mathrm{2}}{{x}}\right)^{{r}} \\ $$$$\mathrm{4}{c}_{{r}} ×{x}^{\mathrm{8}−\mathrm{2}{r}} ×\frac{\mathrm{2}^{{r}} }{{x}^{{r}} } \\ $$$${hence}\:{x}^{\mathrm{8}−\mathrm{2}{r}−{r}} ={x}^{\mathrm{2}} \\ $$$$−\mathrm{3}{r}=\mathrm{2}−\mathrm{8}\:\:\:\rightarrow−\mathrm{3}{r}=−\mathrm{6} \\ $$$${r}=\mathrm{2} \\ $$$$\mathrm{4}{c}_{\mathrm{2}} ×{x}^{\mathrm{8}−\mathrm{3}×\mathrm{2}} ×\mathrm{2}^{\mathrm{2}} \rightarrow\:\:\:\frac{\mathrm{4}!}{\mathrm{2}!\mathrm{2}!}×\mathrm{4}×{x}^{\mathrm{2}} \\ $$$${so}\:{the}\:{coefficient}\:{is}\:\rightarrow\mathrm{6}×\mathrm{4}=\mathrm{24} \\ $$

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