The coefficient of $x^{12}\: \text{in}\: (x^{3}+x^{4}+x^{5}+x^{6}+\dots)^{3}$
$\implies(x^{3}+x^{4}+x^{5}+x^{6}+\dots)^{3}$
$\implies[x^{3}(1+x+x^{2}+x^{3}+x^{4}+\dots)]^{3}$
$\implies[x^{3}(1+x+x^{2}+x^{3}+x^{4}+\dots)]^{3}$
$\implies x^{9}[1+ x+x^{2}+x^{3}+x^{4}+\dots]^{3}\rightarrow(1)$
Let generating sequence $(1,1,1,1,1,1,1\dots)$
write the generating function:
$G(x) = 1+x+x^{2} + x^{3} + x^{4} +\dots$
$G(x) = \dfrac{1}{(1-x)} \:\: [\text{ Sum of infinite series}]$
Put this value in equation $(1),$ and get
$\implies$ $x^{9}\left[\dfrac{1}{(1-x)}\right]^{3}$
$\implies$ $x^{9}.\dfrac{1}{(1-x)^{3}}$
$\implies$ $x^{9}.(1-x)^{-3}$
Apply Binomial Theorem
- $(a+b)^{n}=\sum_{k=0}^{n}\binom{n}{k}(a)^{n-k}.(b)^{k}$
- $(a+b)^{-n}=\sum_{k=0}^{\infty}\binom{-n}{k}(a)^{-n-k}.(b)^{k}$
- $(a+b)^{-n}=\sum_{k=0}^{\infty}\binom{n+k-1}{k}(-1)^{k}(a)^{-n-k}.(b)^{k}$
Now i find the $(1-x)^{-3}=\sum_{k=0}^{\infty}\binom{3+k-1}{k}(-1)^{k}(1)^{-3-k}.(-x)^{k}$
$\implies$$x^{9}.(1-x)^{-3}= x^{9}.\sum_{k=0}^{\infty}\binom{3+k-1}{k}(-1)^{k}.(-x)^{k}$
Put $k =3$,we get $x^{12},$
$\implies x^{9}.\binom{3+3-1}{3}(-1)^{3}.(-x)^{3}$
$\implies x^{9}.\binom{5}{3}(-1)^{3}.(-1)^{3}.(x)^{3}$
$\implies x^{9}.\binom{5}{3}(-1).(-1).(x)^{3}$
$\implies x^{9}.\binom{5}{3}.x^{3}$
$\implies \binom{5}{3}.x^{3}x^{9}.$
Coefficient of $x^{12}\: \text{is} : \binom{5}{3}$
$\implies \dfrac{5!}{(5-3)!.3!}$
$\implies \dfrac{5!}{2!.3!}$
$\implies \dfrac{5.4.3!}{2!.3!}$
$\implies \dfrac{5.4}{2.1 }$
$\implies10$
So, the correct answer is $10.$