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Thread: 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 $\displaystyle x=1+a+a^2+...$ and $\displaystyle y=1+b+b^2$ where |a| <1 and |b| < 1. Prove that $\displaystyle 1 + ab + a^2b^2 + ...= \frac{xy}{x+y-1}$

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

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

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

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

    If I substitute 1 and 2 in 4,i get

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

    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
    409

    You're so close...

    $\displaystyle

    \frac{1-(ab)^n}{1-ab}
    $
    $\displaystyle \frac{(1-a^n)(1-b^n)}{(1-a^n)(1-b)+(1-b^n)(1-a)-(1-a)(1-b)}$
    You're so close! Now, find the limit of both sides as n goes to infinity. Since |a|<1 and |b|<1, $\displaystyle a^n$, $\displaystyle b^n$, and $\displaystyle (ab)^n$ 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 $\displaystyle x=1+a+a^2+...+a^n$ ??? If infinite, then your equations should read:

    $\displaystyle

    x = \frac{1}{1-a} [1]
    $

    $\displaystyle

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