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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 (8.8k points)  
edited by | 1.5k views

6 Answers

+17 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 $10$

 

answered by Veteran (52.2k points)  
selected 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
+4 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 (49.6k 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)
+3 votes
Answer is 10
answered by Junior (551 points)  
I got 55
+3 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 (14.8k 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 :)
+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 (1.3k 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 (911 points)  
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