Menu Close

what-is-coefficient-of-t-3-in-the-expanssion-1-t-6-1-t-3-

Tinku Tara 0 Comments
Pin




Question Number 87074 by jagoll last updated on 02/Apr/20
what is coefficient of t^3 in the expanssion {((1−t^6 )/(1−t))}^3
$$\mathrm{what}\:\mathrm{is}\:\mathrm{coefficient}\:\mathrm{of}\:\mathrm{t}^{\mathrm{3}} \\ $$$$\mathrm{in}\:\mathrm{the}\:\mathrm{expanssion}\:\left\{\frac{\mathrm{1}−\mathrm{t}^{\mathrm{6}} }{\mathrm{1}−\mathrm{t}}\right\}^{\mathrm{3}} \: \\ $$
Commented by mr W last updated on 02/Apr/20
t^2 : C_2 ^4 =6 t^3 : C_2 ^5 =10 t^4 : C_2 ^6 =15 t^5 : C_2 ^7 =21 t^6 : C_2 ^8 −3=25
$${t}^{\mathrm{2}} :\:{C}_{\mathrm{2}} ^{\mathrm{4}} =\mathrm{6} \\ $$$${t}^{\mathrm{3}} :\:{C}_{\mathrm{2}} ^{\mathrm{5}} =\mathrm{10} \\ $$$${t}^{\mathrm{4}} :\:{C}_{\mathrm{2}} ^{\mathrm{6}} =\mathrm{15} \\ $$$${t}^{\mathrm{5}} :\:{C}_{\mathrm{2}} ^{\mathrm{7}} =\mathrm{21} \\ $$$${t}^{\mathrm{6}} :\:{C}_{\mathrm{2}} ^{\mathrm{8}} −\mathrm{3}=\mathrm{25} \\ $$
Commented by mr W last updated on 02/Apr/20
t^(10) : C_2 ^(12) −3×15=21
$${t}^{\mathrm{10}} :\:{C}_{\mathrm{2}} ^{\mathrm{12}} −\mathrm{3}×\mathrm{15}=\mathrm{21} \\ $$
Commented by jagoll last updated on 03/Apr/20
sir how get C_2 ^5 ?
$$\mathrm{sir}\:\mathrm{how}\:\mathrm{get}\:\mathrm{C}_{\mathrm{2}} ^{\mathrm{5}} \:? \\ $$
Commented by mr W last updated on 03/Apr/20
(1−t^6 )^3 has 1 or terms with at least t^6 so we only need to see (1−t)^(−3) . the coef. of its t^k is C_(3−1) ^(k+3−1) =C_2 ^(k+2) , therefore coef. of t^3 is C_2 ^(3+2) =C_2 ^5 .
$$\left(\mathrm{1}−{t}^{\mathrm{6}} \right)^{\mathrm{3}} \:{has}\:\mathrm{1}\:{or}\:{terms}\:{with}\:{at}\:{least}\:{t}^{\mathrm{6}} \\ $$$${so}\:{we}\:{only}\:{need}\:{to}\:{see}\:\left(\mathrm{1}−{t}\right)^{−\mathrm{3}} .\:{the} \\ $$$${coef}.\:{of}\:{its}\:{t}^{{k}} \:{is}\:{C}_{\mathrm{3}−\mathrm{1}} ^{{k}+\mathrm{3}−\mathrm{1}} ={C}_{\mathrm{2}} ^{{k}+\mathrm{2}} ,\:{therefore} \\ $$$${coef}.\:{of}\:{t}^{\mathrm{3}} \:{is}\:{C}_{\mathrm{2}} ^{\mathrm{3}+\mathrm{2}} ={C}_{\mathrm{2}} ^{\mathrm{5}} . \\ $$
Answered by malwaan last updated on 02/Apr/20
{((1−t^6 )/(1−t))}^3 = (1+t+t^2 +t^3 +t^4 +t^5 )^3 = ⇒ ( t^3 + 3t^3 + 6t^3 )=10t^3 (terms contining t^3 only) so the coefficient of t^3 is 10
$$\left\{\frac{\mathrm{1}−\boldsymbol{{t}}^{\mathrm{6}} }{\mathrm{1}−\boldsymbol{{t}}}\right\}^{\mathrm{3}} = \\ $$$$\left(\mathrm{1}+\boldsymbol{{t}}+\boldsymbol{{t}}^{\mathrm{2}} +\boldsymbol{{t}}^{\mathrm{3}} +\boldsymbol{{t}}^{\mathrm{4}} +\boldsymbol{{t}}^{\mathrm{5}} \right)^{\mathrm{3}} \:= \\ $$$$\Rightarrow\:\left(\:\boldsymbol{{t}}^{\mathrm{3}} \:+\:\mathrm{3}\boldsymbol{{t}}^{\mathrm{3}} \:+\:\mathrm{6}\boldsymbol{{t}}^{\mathrm{3}} \:\right)=\mathrm{10}\boldsymbol{{t}}^{\mathrm{3}} \\ $$$$\left(\boldsymbol{{terms}}\:\boldsymbol{{contining}}\:\boldsymbol{{t}}^{\mathrm{3}} \:\boldsymbol{{only}}\right) \\ $$$$\boldsymbol{{so}}\:\:\boldsymbol{{the}}\:\boldsymbol{{coefficient}}\:\boldsymbol{{of}}\:\boldsymbol{{t}}^{\mathrm{3}} \\ $$$$\boldsymbol{{is}}\:\mathrm{10} \\ $$$$ \\ $$

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *