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Comparing exponents in an equation.

  1. Nov 2, 2013 #1
    an + bn/ an-1 + bn-1 = (a+b)/2
    can we say n=1 by comparing exponents?

    is there any other solution of it?
     
    Last edited: Nov 2, 2013
  2. jcsd
  3. Nov 2, 2013 #2

    Simon Bridge

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    $$ \frac{a^n + b^n}{a^{n-1}+b^{n-1}}=\frac{a+b}{2}$$ rearranging: $$a^n+b^n = ab^{n-1} + ba^{n-1}$$

    By inspection, n=1 will satisfy the relation for all real a and b.
    Are there any other examples?

    ##n=2##

    ##a^2+b^2=2ab \Rightarrow (a-b)^2=0##
    ... so the relation with n=2 describes a curve (a parabola).

    So it kinda depends on what counts as an answer.
    (The problem is under-specified.)
     
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