@smsubham I didn't get "Terms beyond x^6 we cannot use to get x^12 here.".

and why are you using three terms? Why not (x^3)*(x^9) to get (x^12)?

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51 votes

@smsubham I didn't get "Terms beyond x^6 we cannot use to get x^12 here.".

and why are you using three terms? Why not (x^3)*(x^9) to get (x^12)?

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67 votes

Best answer

we will get $x^{12}$ as

- $(x^4)^3$ having coefficient $^3C_0=1$
- $(x^3)^2(x^6)$ having coefficient $^3C_1=3$
- $(x^3)(x^4)(x^5)$ 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 \\ $

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51 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.

19 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 $10.$

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for proof, you can check https://gateoverflow.in/205464/extended-binomial-coefficients

It is also given in rosen.

2

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

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

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