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Question Number 83166 by john santu last updated on 28/Feb/20
what is coefficient x^(5 ) in expansion (1+x^2 )^5 ×(1+x)^4
$$\mathrm{what}\:\mathrm{is}\:\mathrm{coefficient}\:\mathrm{x}^{\mathrm{5}\:} \mathrm{in}\:\mathrm{expansion} \\ $$$$\left(\mathrm{1}+\mathrm{x}^{\mathrm{2}} \right)^{\mathrm{5}} ×\left(\mathrm{1}+\mathrm{x}\right)^{\mathrm{4}} \: \\ $$
Commented by mr W last updated on 28/Feb/20
Σ_(k=0) ^5 C_k ^5 x^(2k) Σ_(r=0) ^4 C_r ^4 x^r 2k+r=5 ⇒k=1, r=3 ⇒C_1 ^5 C_3 ^4 ⇒k=2, r=1 ⇒C_2 ^5 C_1 ^4 C_1 ^5 C_3 ^4 +C_2 ^5 C_1 ^4 =20+40=60 ⇒ answer
$$\underset{{k}=\mathrm{0}} {\overset{\mathrm{5}} {\sum}}{C}_{{k}} ^{\mathrm{5}} {x}^{\mathrm{2}{k}} \underset{{r}=\mathrm{0}} {\overset{\mathrm{4}} {\sum}}{C}_{{r}} ^{\mathrm{4}} {x}^{{r}} \\ $$$$\mathrm{2}{k}+{r}=\mathrm{5} \\ $$$$\Rightarrow{k}=\mathrm{1},\:{r}=\mathrm{3}\:\Rightarrow{C}_{\mathrm{1}} ^{\mathrm{5}} {C}_{\mathrm{3}} ^{\mathrm{4}} \\ $$$$\Rightarrow{k}=\mathrm{2},\:{r}=\mathrm{1}\:\Rightarrow{C}_{\mathrm{2}} ^{\mathrm{5}} {C}_{\mathrm{1}} ^{\mathrm{4}} \\ $$$${C}_{\mathrm{1}} ^{\mathrm{5}} {C}_{\mathrm{3}} ^{\mathrm{4}} +{C}_{\mathrm{2}} ^{\mathrm{5}} {C}_{\mathrm{1}} ^{\mathrm{4}} =\mathrm{20}+\mathrm{40}=\mathrm{60}\:\Rightarrow\:{answer} \\ $$
Commented by john santu last updated on 28/Feb/20
thank you
$$\mathrm{thank}\:\mathrm{you} \\ $$
Answered by john santu last updated on 28/Feb/20
Answered by TANMAY PANACEA last updated on 28/Feb/20
(1+5c_1 x^2 +5c_2 x^4 +5c_3 x^6 +5c_4 x^8 +x^(10) )(1+4c_1 x+4_ c_2 x^2 +4c_3 x^3 +x^4 ) 5c_1 x^2 ×4c_3 x^3 +5c_2 x^4 ×4c_1 x =x^5 (5×4+10×4) =60x^5
$$\left(\mathrm{1}+\mathrm{5}{c}_{\mathrm{1}} {x}^{\mathrm{2}} +\mathrm{5}{c}_{\mathrm{2}} {x}^{\mathrm{4}} +\mathrm{5}{c}_{\mathrm{3}} {x}^{\mathrm{6}} +\mathrm{5}{c}_{\mathrm{4}} {x}^{\mathrm{8}} +{x}^{\mathrm{10}} \right)\left(\mathrm{1}+\mathrm{4}{c}_{\mathrm{1}} {x}+\mathrm{4}_{} {c}_{\mathrm{2}} {x}^{\mathrm{2}} +\mathrm{4}{c}_{\mathrm{3}} {x}^{\mathrm{3}} +{x}^{\mathrm{4}} \right) \\ $$$$\mathrm{5}{c}_{\mathrm{1}} {x}^{\mathrm{2}} ×\mathrm{4}{c}_{\mathrm{3}} {x}^{\mathrm{3}} +\mathrm{5}{c}_{\mathrm{2}} {x}^{\mathrm{4}} ×\mathrm{4}{c}_{\mathrm{1}} {x} \\ $$$$={x}^{\mathrm{5}} \left(\mathrm{5}×\mathrm{4}+\mathrm{10}×\mathrm{4}\right) \\ $$$$=\mathrm{60}{x}^{\mathrm{5}} \\ $$
Commented by john santu last updated on 28/Feb/20
thank you
$$\mathrm{thank}\:\mathrm{you} \\ $$
Commented by TANMAY PANACEA last updated on 28/Feb/20
most welcome
$${most}\:{welcome} \\ $$

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