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Math Help - Mathematical Induction

  1. #1
    Member maybeline9216's Avatar
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    Mathematical Induction

    For qn 20a, can u show me how [1/(sqrrt2 +1)] is equal to (sqrt 2 -1)
    by doing workings and nt using decimals.

    And pls teach me qn21(i) the hence part.

    Thks alot!!!
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  2. #2
    MHF Contributor Amer's Avatar
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    Quote Originally Posted by maybeline9216 View Post
    For qn 20a, can u show me how [1/(sqrrt2 +1)] is equal to (sqrt 2 -1)
    by doing workings and nt using decimals.

    And pls teach me qn21(i) the hence part.

    Thks alot!!!
    ok we want to prove that

    \sum_{r=1}^{n} \frac{1}{\sqrt{r+1}+\sqrt{r}}=\sqrt{n+1}-1

    let
    P_k:\sum_{r=1}^{k}\frac{1}{\sqrt{r+1}+\sqrt{r}} = \sqrt{k+1}-1

    lats see if p_k true or not

    P_1 :\frac{1}{\sqrt{2}+1} =\sqrt{2}-1

    \left(\frac{1}{\sqrt{2}+1}\right)\left(\frac{\sqrt  {2}-1}{\sqrt{2}-1}\right) = \frac{\sqrt{2}-1}{2-1} = \sqrt{2}-1
    so P1 is true suppose that Pk is true try P(k+1)


    P_{k+1} : \sum_{r=1}^{k+1}\frac{1}{\sqrt{r+1}+\sqrt{r}} = \sqrt{(k+1)+1}-1=\sqrt{k+2}-1

    the left side

    \sum_{r=1}^{k+1}\frac{1}{\sqrt{r+1}+\sqrt{r}}=\fra  c{1}{\sqrt{k+2}+\sqrt{k+1}}+\sum_{r=1}^{k}\frac{1}  {\sqrt{r+1}+\sqrt{r}}

    but the second term equal  \sqrt{k+1}-1

    \frac{1}{\sqrt{k+2}+\sqrt{k+1}}+\sqrt{k+1}-1

    \left(\frac{1}{\sqrt{k+2}+\sqrt{k+1}}\right)\left(  \frac{\sqrt{k+2}-\sqrt{k+1}}{\sqrt{k+2}-\sqrt{k+1}}\right)+\sqrt{k+1}-1

    \Rightarrow \frac{\sqrt{k+2}-\sqrt{k+1}}{k+2-k-1}+\sqrt{k+1}-1

    \sqrt{k+2}-\sqrt{k+1}+\sqrt{k+1}-1 =\sqrt{k+2}-1

    so P_{k+1} is true we prove it by induction
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