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Math Help - induction problem

  1. #1
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    induction problem

    I have to prove that for all integers n>= 1

    1/3 = (1+3)/(5+7) = (1+3+5)/(7+9+11) = ...  =  1+3+.....+
    (1+3+...+(2n-1))/(2n+1)+....+(4n-1)


    don't know how to go about it.
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  2. #2
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    Hello, NidhiS


    The sum of the first n terms of an arithmetic series is given by:

    . . S_n \;=\;\frac{n}{2}[2a + (n-1)d] . . . where: . \begin{Bmatrix}n &=& \text{no. of terms} \\ a &=& \text{first term} \\ d&=&\text{common diff.} \end{Bmatrix}


    Prove for all integers n \geq 1\!:\;\;

    \frac{1}{3} \;=\;\frac{1+3}{5+7} \;=\; \frac{1+3+5}{7+9+11}) \;=\; \hdots \;=\;\frac{1+3+\hdots + (2n-1)}{(2n+1)+(2n+3)+\hdots+(4n-1)}

    The numerator is: . 1 + 3 + 5 + \hdots + (2k-1)

    This is an arithmetic series with: .  a = 1,\;d = 2,\;n = k

    . . Its sum is: . N \;=\;\frac{k}{2}[2(1) + (k-1)2] \;=\;k^2


    The denominator is: . (2k+1)+(2k+3) + \hdots + (4k-1)

    This is an arithmetic series with: . a = (2k+1),\;d = 2,\;n = k

    . . Its sum is: . D \;=\;\frac{k}{2}[2(2k+1) +(k-1)2] \;=\;3k^2


    Therefore: . \frac{N}{D} \;=\;\frac{k^2}{3k^2} \;=\;\frac{1}{3}

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