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Math Help - GP Series Problem...stuck

  1. #1
    Newbie
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    Apr 2009
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    chennai,india
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    GP Series Problem...stuck

    I am trying to solve the problem and but i am stuck.. Can someone please have look and help out...
    Problem
    if x=1+a+a^2+... and  y=1+b+b^2 where |a| <1 and |b| < 1. Prove that  1 + ab + a^2b^2 + ...= \frac{xy}{x+y-1}

    Solution
    x=1+a+a^2+...
    GP Series
    x = \frac{1-a^n}{1-a}   [1]

    Similarly
    y=\frac{1-b^n}{1-b} [2]

    taking LHS of the equation
     1 + ab + a^2b^2 + ...
    using GP
     \frac{1-(ab)^n}{1-ab}  [3]

    RHS of the equation
    \frac{xy}{x+y-1}  [4]

    If I substitute 1 and 2 in 4,i get

    \frac{(1-a^n)(1-b^n)}{(1-a^n)(1-b)+(1-b^n)(1-a)-(1-a)(1-b)}

    this is no where near to equ 4. My approach might be wrong. please suggest if possible...
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  2. #2
    Senior Member
    Joined
    Apr 2009
    From
    Atlanta, GA
    Posts
    408

    You're so close...

    <br /> <br />
\frac{1-(ab)^n}{1-ab}<br />
    \frac{(1-a^n)(1-b^n)}{(1-a^n)(1-b)+(1-b^n)(1-a)-(1-a)(1-b)}
    You're so close! Now, find the limit of both sides as n goes to infinity. Since |a|<1 and |b|<1, a^n, b^n, and (ab)^n go to zero. Go ahead and make these substitutions and both the LHS and RHS will agree!

    *In the original problem, are you required to prove that this equality holds for a terminating sequence x=1+a+a^2+...+a^n ??? If infinite, then your equations should read:

    <br /> <br />
x = \frac{1}{1-a}   [1]<br />

    <br /> <br />
y = \frac{1}{1-b}   [2]<br />
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  3. #3
    Newbie
    Joined
    Apr 2009
    From
    chennai,india
    Posts
    9
    Thank you.
    No they are not terminating sequence... They are infinite sequence.. I assumed it wrongly... Thanks for correcting me...
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