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Question Number 23700 by Tinkutara last updated on 04/Nov/17

If (1 − x^3 )^n  = Σ_(r=0) ^n a_r x^r (1 − x)^(3n−2r) , then  the value of a_r , where n ∈ N is  (1)^n C_r ∙3^r   (2)^n C_(3r)   (3)^n C_(r−1) 2^(r−1)   (4)^n C_r 2^r

$$\mathrm{If}\:\left(\mathrm{1}\:−\:{x}^{\mathrm{3}} \right)^{{n}} \:=\:\underset{{r}=\mathrm{0}} {\overset{{n}} {\sum}}{a}_{{r}} {x}^{{r}} \left(\mathrm{1}\:−\:{x}\right)^{\mathrm{3}{n}−\mathrm{2}{r}} ,\:\mathrm{then} \\ $$$$\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:{a}_{{r}} ,\:\mathrm{where}\:{n}\:\in\:{N}\:\mathrm{is} \\ $$$$\left(\mathrm{1}\right)\:^{{n}} {C}_{{r}} \centerdot\mathrm{3}^{{r}} \\ $$$$\left(\mathrm{2}\right)\:^{{n}} {C}_{\mathrm{3}{r}} \\ $$$$\left(\mathrm{3}\right)\:^{{n}} {C}_{{r}−\mathrm{1}} \mathrm{2}^{{r}−\mathrm{1}} \\ $$$$\left(\mathrm{4}\right)\:^{{n}} {C}_{{r}} \mathrm{2}^{{r}} \\ $$

Commented by ajfour last updated on 04/Nov/17

If n=1  1−x^3 =a_0 (1−x)^3 +a_1 x(1−x)  ⇒ a_0 =1  and  a_1 =3  if we check options with n=1  (1) gives a_0 =1, a_1 =3  (2) a_0 =1, a_1 =!  (3) a_0 =! , a_1 =1  (4) a_0 =1 , a_1 =2  So only option (1) qualifies..

$${If}\:{n}=\mathrm{1} \\ $$$$\mathrm{1}−{x}^{\mathrm{3}} ={a}_{\mathrm{0}} \left(\mathrm{1}−{x}\right)^{\mathrm{3}} +{a}_{\mathrm{1}} {x}\left(\mathrm{1}−{x}\right) \\ $$$$\Rightarrow\:{a}_{\mathrm{0}} =\mathrm{1}\:\:{and}\:\:{a}_{\mathrm{1}} =\mathrm{3} \\ $$$${if}\:{we}\:{check}\:{options}\:{with}\:{n}=\mathrm{1} \\ $$$$\left(\mathrm{1}\right)\:{gives}\:{a}_{\mathrm{0}} =\mathrm{1},\:{a}_{\mathrm{1}} =\mathrm{3} \\ $$$$\left(\mathrm{2}\right)\:{a}_{\mathrm{0}} =\mathrm{1},\:{a}_{\mathrm{1}} =! \\ $$$$\left(\mathrm{3}\right)\:{a}_{\mathrm{0}} =!\:,\:{a}_{\mathrm{1}} =\mathrm{1} \\ $$$$\left(\mathrm{4}\right)\:{a}_{\mathrm{0}} =\mathrm{1}\:,\:{a}_{\mathrm{1}} =\mathrm{2} \\ $$$${So}\:{only}\:{option}\:\left(\mathrm{1}\right)\:{qualifies}.. \\ $$

Commented by Tinkutara last updated on 05/Nov/17

Not any proof?

$$\mathrm{Not}\:\mathrm{any}\:\mathrm{proof}? \\ $$

Commented by ajfour last updated on 05/Nov/17

i will try ..

$${i}\:{will}\:{try}\:.. \\ $$

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