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The coefficient of $x^{12}$ in $\left(x^{3}+x^{4}+x^{5}+x^{6}+\dots \right)^{3}$ is ___________.
in Combinatory by Loyal (7.2k points)
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+28
$\begin{align*} &\left [ {\color{red}{\bf x^{12}}} \right ] \left ( x^3 + x^4 + x^5 + x^6 + \dots \right )^3 \\ &\left [ {\color{red}{\bf x^{12}}} \right ] \left [ x^3\left ( 1 + x^1 + x^2 + x^3 + \dots \right ) \right ]^3 \\ &\left [ {\color{red}{\bf x^{12}}} \right ] \left [ x^9\left ( 1 + x^1 + x^2 + x^3 + \dots \right )^3 \right ] \\ &\left [ {\color{red}{\bf x^{3}}} \right ] \left [ \left ( 1 + x^1 + x^2 + x^3 + \dots \right )^3 \right ] \\ &\left [ {\color{red}{\bf x^{3}}} \right ] \left [ \left ( \frac{1}{1-x} \right )^3 \right ] \\ &\left [ {\color{red}{\bf x^{3}}} \right ] \left [ \sum_{k=0}^{\infty}\binom{3+k-1}{k}x^k \right ] \\ &\text{Now , put k = 3} \\ &\text{Coefficient of }\left [ {\color{red}{\bf x^{3}}} \right ] = \binom{2+3}{3} = 5C3 = 5C2 = 10 \\ \\ \hline \\ &\left [ {\color{red}{\bf x^{12}}} \right ] \left ( x^3 + x^4 + x^5 + x^6 + \dots \right )^3 \Rightarrow \bf 10 \\ \end{align*}$
+17
This can be solved using common sense.

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

Terms beyond $x^6$  we cannot use to get x^12 here.

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

We can use permutation

a,b,c -> take $x^a$ form term 1, $x^b$ form term 2 and $x^c$ form term 3

3,3,6 -> In 3 p3 / 2 ! ways = 3 ( as 3 is repreated so we have divided by 2!)

3, 4, 5 -> 3 p 3 ways = 6

4 , 4, 4 -> 3 p 3 / 3! ways = 1  ( 4 is repreated thrice so divided by 3!)

So total ways is  3 + 6 + 1= 10 ways.
+1
I am not able to solve generating function questions. How to solve them correctly?
0
Follow Rosen

10 Answers

+48 votes
Best answer

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 \\ $

by Veteran (56.6k points)
edited by
0
Isn't this out of syllabus?
0
could you provide more details, or point to the topic.
+5

Just as we find binomial coefficients 

0
Plz give a detailed answer . How you have calculated the coefficients.

Thanks
0
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
0
please explain 1st method
+1

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

+36 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.

by Boss (23.7k points)
edited by
0
What is the formula for

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

I am felling difficult to get coefficient 6 and 10

plz tell
+19

See 5th row from below.

+2

Trick to remember co-efficient of (1−x)−3 is ...It is  triangular number.

+1
thanks @rajesh for uploading a single page containing all formulas
0
Welcome @Akriti :)
+1
thanks :)
+1
Welcome @srestha :)
0
use Taylor's expansion.
0

(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?

0

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

+14 votes

The coefficient of $x^{12}$ in $(x^{3}+x^{4}+x^{5}+x^{6}+…)^{3} $ is ___________.

$\Rightarrow$  $(x^{3}+x^{4}+x^{5}+x^{6}+…...)^{3}$ 

$\Rightarrow$  $[x^{3}(1+x+x^{2}+x^{3}+x^{4}+…...)]^{3}$ 

$\Rightarrow$  $[x^{3}(1+x+x^{2}+x^{3}+x^{4}+…...)]^{3}$ 

$\Rightarrow$  $x^{9}[1+ x+x^{2}+x^{3}+x^{4}+…]^{3}$ ------------->(1)

Let generating sequence $(1,1,1,1,1,1,1......................)$

write the generating function:

$G(x) = 1+x+x^{2} + x^{3} + x^{4} +............$

$G(x) = \frac{1}{(1-x)}$     [Sum of infinite series]

Put this value in equation(1)

$\Rightarrow$  $x^{9}[\frac{1}{(1-x)}]^{3}$

$\Rightarrow$  $x^{9}.\frac{1}{(1-x)^{3}}$

$\Rightarrow$  $x^{9}.(1-x)^{-3}$

Apply Binomial theorem
 $(a+b)^{n}=\sum_{k=0}^{n}\binom{n}{k}(a)^{n-k}.(b)^{k}$
$\Rightarrow$$(a+b)^{-n}=\sum_{k=0}^{\infty}\binom{-n}{k}(a)^{-n-k}.(b)^{k}$
$\Rightarrow$$(a+b)^{-n}=\sum_{k=0}^{\infty}\binom{n+k-1}{k}(-1)^{k}(a)^{-n-k}.(b)^{k}$

Now i find the $(1-x)^{-3}=\sum_{k=0}^{\infty}\binom{3+k-1}{k}(-1)^{k}(1)^{-3-k}.(-x)^{k}$

$\Rightarrow$$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 $x^{12},$

$\Rightarrow$ $x^{9}.\binom{3+3-1}{3}(-1)^{3}.(-x)^{3}$

$\Rightarrow$ $x^{9}.\binom{5}{3}(-1)^{3}.(-1)^{3}.(x)^{3}$

$\Rightarrow$ $x^{9}.\binom{5}{3}(-1).(-1).(x)^{3}$

$\Rightarrow$ $x^{9}.\binom{5}{3}.x^{3}$

$\Rightarrow$ $\binom{5}{3}.x^{3}x^{9}.$

Coefficient of $x^{12}\: \text{is} : \binom{5}{3}$

                                  $\Rightarrow \frac{5!}{(5-3)!.3!}$

                                  $\Rightarrow \frac{5!}{2!.3!}$

                                  $\Rightarrow \frac{5.4.3!}{2!.3!}$

                                   $\Rightarrow \frac{5.4}{2.1 }$

                                   $\Rightarrow10$

So, the correct answer  is 10.

by Veteran (53.4k points)
edited by
0
correct ans 10 . nice explaination
0

@royal shubham

Thanks Brother

+1

@Lakshman Patel RJIT    correct the third binomial formula $(a+b)^{-n}$  in place of + it should be -

EDIT    formula is correct  my assumption was wrong 

0
The formula is right.where you want to correction?

Edit:Ok good
0

@Lakshman Patel RJIT 

$(a+b)^{-n}=\sum_{k=0}^\infty(^{-n}C_k)(a^{-n-k})(b^k)$

$=\sum_{k=0}^\infty\left(^{n+k-1}C_{k}\right)(-1)^k(a^{-n-k})(b^k)$

How $^{-n}C_k=^{n+k-1}C_k(-1)^k$  ?

+1

@Verma Ashish

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

It is also given in rosen.

+2
Thanks.. :)

Now i understand..
+11 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
by Veteran (117k points)
edited by
0
@srestha,could'nt understand your soltuion..can you explain it more?
0

We know the formula (a+b)n

                              =n0Ca0bn+n1Ca1bn−1+......

then why you did not take rest terms.you have just taken 3C1* (x3) * (x4 + x5 +..)2.

3C0 * (x3)0 * (x4 + x5 ..) + 3C1* (x3) * (x4 + x5 +..)2. + 3C2* (x3)2 * (x4 + x5 +..)1. +  3C3* (x3) * (x4 + x5 +..)0

0

@Akriti I just have derived more, as I need coefficient of x12

0
sorry,i did not get you..
0

See, for getting coefficient of x12 we need put this formula to see in which sequence I am getting x12

For that I can see x's power.

power of x4,x5,x6 giving x12 , rt?

then find it's coefficient. getting?

+20
@Akriti this might help...

$(x^3 + x^4 + x^5 + x^6 + ... )^3 = (x^9) (1 + x^1 + x^3 + x^4 + x^5 + ...)^3$
From second term we need to find the coefficient of $x^3$

$(1 + x^1 + x^3 + x^4 + x^5 + ...)^3 = (\frac{1}{1-x})^3 = (1-x)^-3 = 1 + 3x + 6x^2 + 10x^3 +...$
Clearly, coefficient of $x^3 is 10$
0
thankyou :-)
0
What is the formula for

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

I am felling difficult to get coefficient 6 and 10

plz tell
+3
summation 0 to infinity (3+r-1 Cr * x^r)
+9

There you go.

+6 votes

= [ x( 1 + x + x+ x+ ....................... ) ]3

= x( 1 + x + x+ x+.......................)3

= x(1 / ( 1 - x ) )3

= x( 1 / ( 1 - x ))

= x( ( 1 - x )-3 )

= x( 1 + (3)x + ( (3*4) / (1*2) ) x+ ( (3*4*5) / (1*2*3) ) x+...................................)

= x+ 3x10 + 6x11 + 10x12 +........................................................

Answer = 10

by (71 points)
+4 votes

[x12](x3 + x4 + x5 +x6 +...)3 = [x3](1 + x + x2 + x3 +x4 +...)= [x3]((1-x)-1)3 =  [x3](1-x)-3

 = -3C3(-1)3 = 5C3(-1)3(-1)=  5C= 10

by Active (1.2k points)
+4 votes

best way of solving these type of question...

by Active (4.4k points)
+2 votes
Answer is 10
by Junior (701 points)
–4
I got 55
Answer:

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