19,679 views
The coefficient of $x^{12}$ in $\left(x^{3}+x^{4}+x^{5}+x^{6}+\dots \right)^{3}$ is ___________.

@ 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)?

So you have to add to 12 using {3,4,6...}

3 3 6

3 4 5

3 5 4

3 6 3

3 7 2 (not possible because min 3 you have to use )

4 3 5

4 4 4

4 5 3

now continue this

5 3 4

5 4 3

6 3 3

So you can see total 10 ways are there, if you understand this you can do in like 50 sec in exam

Using  Stars and Bars.

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:

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

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

we will get x12x12 as

1. (x4)3(x4)3 having coefficient 3C0=13C0=1

2. (x3)2(x6)(x3)2(x6) having coefficient 3C1=33C1=3

3. (x3)(x4)(x5)(x3)(x4)(x5) having coefficient 3C2×2C1=63C2×2C1=6

So it is 10

3C0=13C0=1

3C1=33C1=3...........  how  do ths  @praveen

for generating functions refer this site http://discretetext.oscarlevin.com/dmoi/section-27.html

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

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.

(x^3+x^4+x^5+x^6+...)^3=(x^9)(1+x^1+x^2+x^3+x^4+...)^3

This will be the trick no?

@Rajesh Pradhan sir can you upload that sheet again it's not showing up

edited

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 $10.$

@Verma Ashish

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

It is also given in rosen.

Thanks.. :)

Now i understand..
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
by

What is the formula for

$(1-x)^{-3}$

I am felling difficult to get coefficient 6 and 10

plz tell
summation 0 to infinity (3+r-1 Cr * x^r)

There you go.

1
19,637 views