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Thread: induction problem

  1. #1
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    induction problem

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

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


    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 $\displaystyle n$ terms of an arithmetic series is given by:

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


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

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

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

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

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


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

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

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


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

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