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Question Number 88882 by Zainal Arifin last updated on 13/Apr/20

The coefficient of x^3 in ((√x^5 ) +(3/(√x^3 )))^6 is

$$\mathrm{The}\:\mathrm{coefficient}\:\mathrm{of}\:\:{x}^{\mathrm{3}} \:\mathrm{in}\:\left(\sqrt{{x}^{\mathrm{5}} }\:+\frac{\mathrm{3}}{\sqrt{{x}^{\mathrm{3}} }}\right)^{\mathrm{6}} \:\:\mathrm{is} \\ $$

Commented by mathmax by abdo last updated on 13/Apr/20

we have ((√x^5 )+(3/(√x^3 )))^6 =(x^(1/5) +3x^(−(3/2)) )^6 =Σ_(k=0) ^6 C_6 ^k (3x^(−(3/2)) )^k (x^(5/2) )^(6−k) =Σ_(k=0) ^6 C_6 ^k 3^k x^(−((3k)/2)) x^((30−5k)/2) =Σ_(k=0) ^6 C_6 ^k 3^k x^((−3k+30−5k)/2) we get the coefficient of x^3 when ((−8k+30)/2) =3 ⇒ −8k+30 =6 ⇒8k =24 ⇒k=3 and the coefficient is a =C_6 ^3 3^3 =27×((6!)/(3!×3!)) =((27.6.5.4.3!)/(3!.3!)) =27.5.4 =27×20 =540

$${we}\:{have}\:\:\left(\sqrt{{x}^{\mathrm{5}} }+\frac{\mathrm{3}}{\sqrt{{x}^{\mathrm{3}} }}\right)^{\mathrm{6}} \:=\left({x}^{\frac{\mathrm{1}}{\mathrm{5}}} \:+\mathrm{3}{x}^{−\frac{\mathrm{3}}{\mathrm{2}}} \right)^{\mathrm{6}} \\ $$$$=\sum_{{k}=\mathrm{0}} ^{\mathrm{6}} \:{C}_{\mathrm{6}} ^{{k}} \left(\mathrm{3}{x}^{−\frac{\mathrm{3}}{\mathrm{2}}} \right)^{{k}} \left({x}^{\frac{\mathrm{5}}{\mathrm{2}}} \right)^{\mathrm{6}−{k}} \\ $$$$=\sum_{{k}=\mathrm{0}} ^{\mathrm{6}} \:{C}_{\mathrm{6}} ^{{k}} \:\mathrm{3}^{{k}} \:{x}^{−\frac{\mathrm{3}{k}}{\mathrm{2}}} \:{x}^{\frac{\mathrm{30}−\mathrm{5}{k}}{\mathrm{2}}} \:=\sum_{{k}=\mathrm{0}} ^{\mathrm{6}} \:{C}_{\mathrm{6}} ^{{k}} \:\mathrm{3}^{{k}} \:{x}^{\frac{−\mathrm{3}{k}+\mathrm{30}−\mathrm{5}{k}}{\mathrm{2}}} \\ $$$${we}\:{get}\:{the}\:{coefficient}\:{of}\:{x}^{\mathrm{3}} \:{when}\:\frac{−\mathrm{8}{k}+\mathrm{30}}{\mathrm{2}}\:=\mathrm{3}\:\Rightarrow \\ $$$$−\mathrm{8}{k}+\mathrm{30}\:=\mathrm{6}\:\Rightarrow\mathrm{8}{k}\:=\mathrm{24}\:\Rightarrow{k}=\mathrm{3}\:\:{and}\:{the}\:{coefficient}\:{is} \\ $$$${a}\:={C}_{\mathrm{6}} ^{\mathrm{3}} \:\mathrm{3}^{\mathrm{3}} \:=\mathrm{27}×\frac{\mathrm{6}!}{\mathrm{3}!×\mathrm{3}!}\:=\frac{\mathrm{27}.\mathrm{6}.\mathrm{5}.\mathrm{4}.\mathrm{3}!}{\mathrm{3}!.\mathrm{3}!}\:=\mathrm{27}.\mathrm{5}.\mathrm{4}\:=\mathrm{27}×\mathrm{20}\:=\mathrm{540} \\ $$

Answered by TANMAY PANACEA. last updated on 13/Apr/20

let (r+1)th term contains x^3 6C_r .((√x^5 ) )^(6−r) .((3/(√x^3 )))^r ((6!)/(r!(6−r)!)).(x^(5/2) )^(6−r) .3^r .(1/((x^(3/2) )^r )) ((6!)/(r!(6−r)!)).(x)^((30−5r)/2) .3^r .(1/x^((3r)/2) ) ((6!×3^r )/(r!(6−r)!))×x^(((30−5r)/2)−((3r)/2)) ((6!×3^r )/(r!(6−r)!))×x^((30−8r)/2) so x^(15−4r) =x^3 −4r=−12 so r=3 ((6!×3^3 )/(3!3!))×x^3 required coefficient=((6×5×4×27)/6)=540

$${let}\:\left({r}+\mathrm{1}\right){th}\:{term}\:{contains}\:{x}^{\mathrm{3}} \\ $$$$\mathrm{6}{C}_{{r}} .\left(\sqrt{{x}^{\mathrm{5}} }\:\right)^{\mathrm{6}−{r}} .\left(\frac{\mathrm{3}}{\sqrt{{x}^{\mathrm{3}} }}\right)^{{r}} \\ $$$$\frac{\mathrm{6}!}{{r}!\left(\mathrm{6}−{r}\right)!}.\left({x}^{\frac{\mathrm{5}}{\mathrm{2}}} \right)^{\mathrm{6}−{r}} .\mathrm{3}^{{r}} .\frac{\mathrm{1}}{\left({x}^{\frac{\mathrm{3}}{\mathrm{2}}} \right)^{{r}} } \\ $$$$\frac{\mathrm{6}!}{{r}!\left(\mathrm{6}−{r}\right)!}.\left({x}\right)^{\frac{\mathrm{30}−\mathrm{5}{r}}{\mathrm{2}}} .\mathrm{3}^{{r}} .\frac{\mathrm{1}}{{x}^{\frac{\mathrm{3}{r}}{\mathrm{2}}} } \\ $$$$\frac{\mathrm{6}!×\mathrm{3}^{{r}} }{{r}!\left(\mathrm{6}−{r}\right)!}×{x}^{\frac{\mathrm{30}−\mathrm{5}{r}}{\mathrm{2}}−\frac{\mathrm{3}{r}}{\mathrm{2}}} \\ $$$$\frac{\mathrm{6}!×\mathrm{3}^{{r}} }{{r}!\left(\mathrm{6}−{r}\right)!}×{x}^{\frac{\mathrm{30}−\mathrm{8}{r}}{\mathrm{2}}} \\ $$$${so}\:\:{x}^{\mathrm{15}−\mathrm{4}{r}} ={x}^{\mathrm{3}} \\ $$$$−\mathrm{4}{r}=−\mathrm{12}\:\:\:{so}\:{r}=\mathrm{3} \\ $$$$\frac{\mathrm{6}!×\mathrm{3}^{\mathrm{3}} }{\mathrm{3}!\mathrm{3}!}×{x}^{\mathrm{3}} \\ $$$${required}\:{coefficient}=\frac{\mathrm{6}×\mathrm{5}×\mathrm{4}×\mathrm{27}}{\mathrm{6}}=\mathrm{540} \\ $$

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