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What is the coefficient of $ab^2c^3$ in $(a + 2b + 3c)^6$?

 May 7, 2015

Best Answer 

 #2
avatar+99122 
+15

I don't remember ever doing one like this before   

I will give it a shot :)

 

$$\\(a+2b+3c)^6\\\\
=(a+[2b+3c])^6\\\\
=\displaystyle\sum_{r=0}^{6}^6C_r(a^r)(2b+3c)^{6-r}\\\\
$Now the degree of a must be 1 so I am only interested in terms where $ r=1\\\\
^6C_1(a^1)(2b+3c)^{5}\\\\
(6a)*[(2b+3c)^{5}]\\\\$$

 

$$\\$The general term for this is $\\\\
6a*[ ^5C_k*(2b)^k(3c)^{5-k}]\\\\
$The power of b must be 2 so k=2\\\\
6a*[ ^5C_2*(2b)^2(3c)^{3}]\\\\
6a*[ 10*4b^2*27c^3]\\\\
6a*[ 10*4b^2*27c^3]\\\\
6x*[1080*b^2c^3]\\\\
6480xb^2c^3\\\\
$So the coefficient of $ bc^2c^3 $ is 6480$$

 

I assume Chris got his answer from Wolfram|Alpha  ?   

Anyway they are the same so we are on a roll.       

 May 7, 2015
 #1
avatar+98044 
+10

6480......

Don't ask me how I know.....it just is.....!!!!!

 

  

 May 7, 2015
 #2
avatar+99122 
+15
Best Answer

I don't remember ever doing one like this before   

I will give it a shot :)

 

$$\\(a+2b+3c)^6\\\\
=(a+[2b+3c])^6\\\\
=\displaystyle\sum_{r=0}^{6}^6C_r(a^r)(2b+3c)^{6-r}\\\\
$Now the degree of a must be 1 so I am only interested in terms where $ r=1\\\\
^6C_1(a^1)(2b+3c)^{5}\\\\
(6a)*[(2b+3c)^{5}]\\\\$$

 

$$\\$The general term for this is $\\\\
6a*[ ^5C_k*(2b)^k(3c)^{5-k}]\\\\
$The power of b must be 2 so k=2\\\\
6a*[ ^5C_2*(2b)^2(3c)^{3}]\\\\
6a*[ 10*4b^2*27c^3]\\\\
6a*[ 10*4b^2*27c^3]\\\\
6x*[1080*b^2c^3]\\\\
6480xb^2c^3\\\\
$So the coefficient of $ bc^2c^3 $ is 6480$$

 

I assume Chris got his answer from Wolfram|Alpha  ?   

Anyway they are the same so we are on a roll.       

Melody May 7, 2015
 #3
avatar+98044 
+10

Yep....I cheated....your method was far more impressive.....!!!!

 This is a GOOD ONE to learn from....!!!!

 

  

 May 7, 2015
 #4
avatar+99122 
+10

Thanks Chris,

I have taken your lead and put it in the 'great answers to learn" from thread   

 May 8, 2015
 #5
avatar+98044 
0

Here's an addendum to this :

 

Let a  = m      and  [ 2b + 3c]  =  n

 

So we are looking for the term in (m + n)^6  where  a = m has a power of 1....and this is given by

 

C(6,1) mn^5  =  6an^5

 

So....we have

 

6a [ 2b + 3c]^5

 

And now we are looking for the term in (2b + 3c)^5  where  [2b]  has a power of 2....and this is given by:

 

C(5, 2) [2b]^2 [3b]^3  =   10 * 4b^2 * 27c^3  =  1080 b^2 c^3

 

Putting this all together, the term we are looking for is given by

 

6a [ 1080 b^2 c^3]  =   6480ab^2c^3 .......   and 6480  is the coefficient

 

 

 

cool cool cool

 Dec 26, 2016

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