SOLUTION: {{{a=1+2x+4x^2+..............}}}, where -1<2x<1, {{{b= 1+3y+9y^2+.............}}}, where -1<3y<1 and 3y+2x=1, prove that ab=a+b

Algebra ->  Equations -> SOLUTION: {{{a=1+2x+4x^2+..............}}}, where -1<2x<1, {{{b= 1+3y+9y^2+.............}}}, where -1<3y<1 and 3y+2x=1, prove that ab=a+b      Log On


   

Question 622156: a=1%2B2x%2B4x%5E2%2B.............., where -1<2x<1, b=+1%2B3y%2B9y%5E2%2B............., where -1<3y<1 and 3y+2x=1, prove that ab=a+b
Answer by Edwin McCravy(20054) About Me  (Show Source):
You can put this solution on YOUR website!
a = 1%2B2x%2B4x%5E2%2B%22...%22 = %282x%29%5E0%2B%282x%29%5E1%2B%282x%29%5E2%2B%22...%22

That's an infinite geometric series with first term t1 = 1,
and r = 2x.  And since -1 < 2x < 1 it converges and we can use the formula:

S%5Binfinity%5D = t%5B1%5D%2F%281-r%29 = 1%2F%281-2x%29


b = 1%2B3y%2B9y%5E2%2B%22...%22 = %283y%29%5E0%2B%283y%29%5E1%2B%283y%29%5E2%2B%22...%22


That's also an infinite geometric series with first term t1 = 3y,
and r = 3y.  And since -1 < 3y < 1 we can use the formula:

S%5Binfinity%5D = t%5B1%5D%2F%281-r%29 = 1%2F%281-3y%29 

So 

a = 1%2F%281-2x%29, b = 1%2F%281-3y%29

ab =  1%2F%281-2x%29·1%2F%281-3y%29 = 1%2F%28%281-2x%29%281-3y%29%29

a + b = 1%2F%281-2x%29+{1/(1-3y)}}} = %28%281-3y%29%2B%281-2x%29%29%2F%28%281-2x%29%281-3y%29%29 =

%281-3y%2B1-2x%29%2F%28%281-2x%29%281-3y%29%29 = %282-3y-2x%29%2F%28%281-2x%29%281-3y%29%29 =

%282-%283y%2B2x%29%29%2F%28%281-2x%29%281-3y%29%29

and we are given that 3y+2x=1, so we replace(3y+2x) by 1

%282-1%29%2F%28%281-2x%29%281-3y%29%29 = 1%2F%28%281-2x%29%281-3y%29%29

So ab and a + b  both equal to 1%2F%28%281-2x%29%281-3y%29%29

Edwin