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The coefficient of $x^{12}$ in $\left(x^{3}+x^{4}+x^{5}+x^{6}+\dots \right)^{3}$ is ___________.
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Best answer
70 votes
70 votes

we will get $x^{12}$ as
 

  1. $(x^4)^3$ having coefficient $^3C_0=1$
  2. $(x^3)^2(x^6)$ having coefficient $^3C_1=3$
  3. $(x^3)(x^4)(x^5)$ having coefficient ${^3C_2} \times {^2C_1} =6$


So it is $10$

Second Method:

$\left [ {\color{red}{\bf x^{12}}} \right ] \left ( x^3 + x^4 + x^5 + x^6 + \dots \right )^3$
$  \left [ {\color{red}{\bf x^{12}}} \right ] \left [ x^3\left ( 1 + x^1 + x^2 + x^3 + \dots \right ) \right ]^3 $
$ \left [ {\color{red}{\bf x^{12}}} \right ] \left [ x^9\left ( 1 + x^1 + x^2 + x^3 + \dots \right )^3 \right ] $
$ \left [ {\color{red}{\bf x^{3}}} \right ] \left [ \left ( 1 + x^1 + x^2 + x^3 + \dots \right )^3 \right ] $
$ \left [ {\color{red}{\bf x^{3}}} \right ] \left [ \left ( \frac{1}{1-x} \right )^3 \right ]$
$ \left [ {\color{red}{\bf x^{3}}} \right ] \left [ \sum_{k=0}^{\infty}\binom{3+k-1}{k}x^k \right ]$
$ \text{Now , put k = 3}$
$ \text{Coefficient of }\left [ {\color{red}{\bf x^{3}}} \right ] = \binom{2+3}{3} = 5C3 = 5C2 = 10 $


$ \left [ {\color{red}{\bf x^{12}}} \right ] \left ( x^3 + x^4 + x^5 + x^6 + \dots \right )^3 \Rightarrow \bf 10 \\ $

edited by
55 votes
55 votes

$(x^3 + x^4 + x^5 + x^6 + ... )^3 = (x^9) (1 + x^1 + x^3 + x^4 + x^5 + ...)^3$  

From second term we need to find the coefficient of $x^3$   bcz ($x^3$ *$x^9$ = x12)

$(1 + x^1 + x^3 + x^4 + x^5 + ...)^3 = (\frac{1}{1-x})^3 = (1-x)^-3 = 1 + 3x + 6x^2 + 10x^3 +...$


Clearly, coefficient of $x^3$ is $10$

Note:- Actually this was ans. from Sudarshana Tripathy which i thought that very easy to understand by anyone. but it was in the one of the comment in one of the ans  here which may not be come under eye of everyone..so i thought that it is worth to put as answer.

edited by
25 votes
25 votes

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 $10.$

edited by
12 votes
12 votes
We know the formula $(a+b)^n$

                              =$_{0}^{n}\textrm{C} a^0 b^n + _{1}^{n}\textrm{C} a^1b^n-1+......$

Similarly,

1) $(_{1}^{3}\textrm{C}) (x^3)^1 (x^4+x^5+x^6+..)^2$

Now for,

$(x^4+x^5+x^6+..)^2$

=$((x^4+x^5)+(x^6+....))^2$

=$(_{0}^{2}\textrm{C}) (x^4+x^5)^0(x^6+..)^2$

 

2) $(x^3+x^4+x^5+x^6+..)^3$

=$((x^3)+(x^4+x^5+x^6+....))^3$

=$(_{0}^{3}\textrm{C}) (x^3)^0(x^4+x^5..)^3$

=$(_{0}^{3}\textrm{C}) (x^3)^0(_{3}^{3}\textrm{C})(x^4)^3(x^5+x^6.....)^0$

3) $(x^3+x^4+x^5+x^6+..)^3 =(_{1}^{3}\textrm{C}) (x^3)^1(_{1}^{2}\textrm{C}) (x^4)^1(_{1}^{1}\textrm{C}) (x^5)^1(x^6+..)^0$

Now addition of coefficients are =(1)+(2)+(3)=10
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