what is coefficient of t^3 in the expanssion {((1−t^6 )/(1−t))}^3


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Question Number 87074 by jagoll last updated on 02/Apr/20
$what is coefficient of t^3 in the expanssion {((1−t^6 )/(1−t))}^3$
$what is coefficient of t^{3}$ $in the expanssion {\frac{1 - t^{6}}{1 - t}}^{3}$
Commented by mr W last updated on 02/Apr/20

$t^{2} : C_{2}^{4} = 6$ $t^{3} : C_{2}^{5} = 10$ $t^{4} : C_{2}^{6} = 15$ $t^{5} : C_{2}^{7} = 21$ $t^{6} : C_{2}^{8} - 3 = 25$
Commented by mr W last updated on 02/Apr/20

$t^{10} : C_{2}^{12} - 3 \times 15 = 21$
Commented by jagoll last updated on 03/Apr/20

$sir how get C_{2}^{5} ?$
Commented by mr W last updated on 03/Apr/20

${(1 - t^{6})}^{3} h a s 1 o r t e r m s w i t h a t l e a s t t^{6}$ $s o w e o n l y n e e d t o s e e {(1 - t)}^{- 3} . t h e$ $c o e f . o f i t s t^{k} i s C_{3 - 1}^{k + 3 - 1} = C_{2}^{k + 2}, t h e r e f o r e$ $c o e f . o f t^{3} i s C_{2}^{3 + 2} = C_{2}^{5} .$
	Answered by malwaan last updated on 02/Apr/20
	${((1−t^6 )/(1−t))}^3 = (1+t+t^2 +t^3 +t^4 +t^5 )^3 = ⇒ ( t^3 + 3t^3 + 6t^3 )=10t^3 (terms contining t^3 only) so the coefficient of t^3 is 10$
	${\frac{1 - t^{6}}{1 - t}}^{3} =$ ${(1 + t + t^{2} + t^{3} + t^{4} + t^{5})}^{3} =$ $\Rightarrow (t^{3} + 3 t^{3} + 6 t^{3}) = 10 t^{3}$ $(t e r m s c o n t i n i n g t^{3} o n l y)$ $s o t h e c o e f f i c i e n t o f t^{3}$ $i s 10$
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