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The coefficient of $x^{12}$ in $\left(x^{3}+x^{4}+x^{5}+x^{6}+\dots \right)^{3}$ is ___________.
asked in Set Theory & Algebra by Boss (8.9k points)  
edited by | 2.7k views
$\begin{align*} &\left [ {\color{red}{\bf x^{12}}} \right ] \left ( x^3 + x^4 + x^5 + x^6 + \dots \right )^3 \\ &\left [ {\color{red}{\bf x^{12}}} \right ] \left [ x^3\left ( 1 + x^1 + x^2 + x^3 + \dots \right ) \right ]^3 \\ &\left [ {\color{red}{\bf x^{12}}} \right ] \left [ x^9\left ( 1 + x^1 + x^2 + x^3 + \dots \right )^3 \right ] \\ &\left [ {\color{red}{\bf x^{3}}} \right ] \left [ \left ( 1 + x^1 + x^2 + x^3 + \dots \right )^3 \right ] \\ &\left [ {\color{red}{\bf x^{3}}} \right ] \left [ \left ( \frac{1}{1-x} \right )^3 \right ] \\ &\left [ {\color{red}{\bf x^{3}}} \right ] \left [ \sum_{k=0}^{\infty}\binom{3+k-1}{k}x^k \right ] \\ &\text{Now , put k = 3} \\ &\text{Coefficient of }\left [ {\color{red}{\bf x^{3}}} \right ] = \binom{2+3}{3} = 5C3 = 5C2 = 10 \\ \\ \hline \\ &\left [ {\color{red}{\bf x^{12}}} \right ] \left ( x^3 + x^4 + x^5 + x^6 + \dots \right )^3 \Rightarrow \bf 10 \\ \end{align*}$

6 Answers

+24 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:

$\begin{align*} &\left [ {\color{red}{\bf x^{12}}} \right ] \left ( x^3 + x^4 + x^5 + x^6 + \dots \right )^3 \\ &\left [ {\color{red}{\bf x^{12}}} \right ] \left [ x^3\left ( 1 + x^1 + x^2 + x^3 + \dots \right ) \right ]^3 \\ &\left [ {\color{red}{\bf x^{12}}} \right ] \left [ x^9\left ( 1 + x^1 + x^2 + x^3 + \dots \right )^3 \right ] \\ &\left [ {\color{red}{\bf x^{3}}} \right ] \left [ \left ( 1 + x^1 + x^2 + x^3 + \dots \right )^3 \right ] \\ &\left [ {\color{red}{\bf x^{3}}} \right ] \left [ \left ( \frac{1}{1-x} \right )^3 \right ] \\ &\left [ {\color{red}{\bf x^{3}}} \right ] \left [ \sum_{k=0}^{\infty}\binom{3+k-1}{k}x^k \right ] \\ &\text{Now , put k = 3} \\ &\text{Coefficient of }\left [ {\color{red}{\bf x^{3}}} \right ] = \binom{2+3}{3} = 5C3 = 5C2 = 10 \\ \\ \hline \\ &\left [ {\color{red}{\bf x^{12}}} \right ] \left ( x^3 + x^4 + x^5 + x^6 + \dots \right )^3 \Rightarrow \bf 10 \\ \end{align*}$
answered by Veteran (52.9k points)  
edited by
Isn't this out of syllabus?
could you provide more details, or point to the topic.

Just as we find binomial coefficients 

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

Thanks
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
+6 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
answered by Veteran (53.7k points)  
edited by
@srestha,could'nt understand your soltuion..can you explain it more?

 

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

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

sorry,i did not get you..

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?

@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 $x^3 is 10$
thankyou :-)
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.

+4 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.

answered by Veteran (16.8k points)  
edited by
What is the formula for

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

I am felling difficult to get coefficient 6 and 10

plz tell

See 5th row from below.

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

thanks @rajesh for uploading a single page containing all formulas
Welcome @Akriti :)
thanks :)
Welcome @srestha :)
+3 votes
Answer is 10
answered by Junior (683 points)  
I got 55
+2 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

answered by Active (1.5k points)  
0 votes

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

now we have to find the coefficient of x^12/x^9 = x^3

this can be found in three ways - 3 OR 1+2 OR 1+1+1 = 3C1+3C1x2C1+3C3 =10

answered by Junior (995 points)  


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