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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
whatiscoefficientoft3intheexpanssion{1t61t}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
t2:C24=6t3:C25=10t4:C26=15t5:C27=21t6:C283=25
Commented by mr W last updated on 02/Apr/20
t^(10) : C_2 ^(12) −3×15=21
t10:C2123×15=21
Commented by jagoll last updated on 03/Apr/20
sir how get C_2 ^5  ?
sirhowgetC25?
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 .
(1t6)3has1ortermswithatleastt6soweonlyneedtosee(1t)3.thecoef.ofitstkisC31k+31=C2k+2,thereforecoef.oft3isC23+2=C25.
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
{1t61t}3=(1+t+t2+t3+t4+t5)3=(t3+3t3+6t3)=10t3(termscontiningt3only)sothecoefficientoft3is10

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