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.
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.
Yep....I cheated....your method was far more impressive.....!!!!
This is a GOOD ONE to learn from....!!!!
Thanks Chris,
I have taken your lead and put it in the 'great answers to learn" from thread
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