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}]\\\\
$Thegeneraltermforthisis$6a∗[5Ck∗(2b)k(3c)5−k]$Thepowerofbmustbe2sok=26a∗[5C2∗(2b)2(3c)3]6a∗[10∗4b2∗27c3]6a∗[10∗4b2∗27c3]6x∗[1080∗b2c3]6480xb2c3$Sothecoefficientof$bc2c3$is6480
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}]\\\\
$Thegeneraltermforthisis$6a∗[5Ck∗(2b)k(3c)5−k]$Thepowerofbmustbe2sok=26a∗[5C2∗(2b)2(3c)3]6a∗[10∗4b2∗27c3]6a∗[10∗4b2∗27c3]6x∗[1080∗b2c3]6480xb2c3$Sothecoefficientof$bc2c3$is6480
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