Product of the Coefficients

[jeremyp1988]

Could anyone help me with these Prod. of the Coefficients problems:

product of the coefficients: (a+b)^3

product of the coefficients: (3a+3b)^2

product of the coefficients: (a+b)^5

[bradp]

http://forum.texasmath.org/viewtopic.ph ... highlight=

[fht]

The coefficient of expanded polynomial of nth degrees follows the (n+1)th row of the Pascal Triangle:
1
1 1
1 2 1 -------- a^2+2ab+b^2
1 3 3 1 -------- a^3+3a^2b+3ab^2+b^3
1 4 6 4 1
1 5 10 10 5 1
............................
Therefore the product of coeff for (x+y)^3 is 1*3*3*1 = 9
for (ax+by)^2 is 2(ab)^3 = 2(3*3)^3 = 1458
for (x+y)^5 is 1*5*10*10*5*1 = 2500

Best Wishes[/b]

[Bumwipe]

[AuSmith]

So, the product of the coefficients of (Aa+Bb)^n is

n=2: 2(AB)^3
n=3: 9(AB)^6
n=4: 96(AB)^10
n=5: 2500(AB)^15

(a+b)^3: 9(1)^6=9
(3a+3b)^2: 2(3*3)^3=1458
(a+b)^5: 2500(1)^15=2500

[Quelloquialism]

[MistaPotta]

Sorry Riley, you were warned.

[fht]

Whatever that was wrong is correct now.
Bumwipe, you are excused.