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51 votes
51 votes
The coefficient of $x^{12}$ in $\left(x^{3}+x^{4}+x^{5}+x^{6}+\dots \right)^{3}$ is ___________.
in Combinatory retagged by
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4 Comments

 @ 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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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

 

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Using  Stars and Bars.

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17 Answers

7 votes
7 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

4 votes
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

4 votes
4 votes

best way of solving these type of question...

3 votes
3 votes

Answer will be 10

we will see why

Answer:

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