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Math Help - Sigma Proof by Induction

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    Sigma Proof by Induction

    Prove 1/1(2)+2/2(3)+...+1/n(n+1)=n/n+1

    I've been using induction to prove this.

    For the base case I got: 1/1(2)=1/2

    Then I go on and get a little more stuck:

    Assume 1/1(2)+2/2(3)+...+1/n(n+1)=n/n+1 for all n in N. Want to show that 1/1(2)+2/2(3)+...+1/n(n+1)+1/(n+1)(n+2)=(n+1)/(n+1)+1
    I simplified it using the assumption

    n/(n+1)+1/(n+1)(n+2)=(n+1)/(n+1)+1
    Then when I tried to simplify I keep getting something that isn't an equality. Have I done something wrong so far?
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    Re: Sigma Proof by Induction

    Quote Originally Posted by renolovexoxo View Post
    Prove 1/1(2)+2/2(3)+...+1/n(n+1)=n/n+1

    I've been using induction to prove this.

    For the base case I got: 1/1(2)=1/2

    Then I go on and get a little more stuck:

    Assume 1/1(2)+2/2(3)+...+1/n(n+1)=n/n+1 for all n in N. Want to show that 1/1(2)+2/2(3)+...+1/n(n+1)+1/(n+1)(n+2)=(n+1)/(n+1)+1
    I simplified it using the assumption

    n/(n+1)+1/(n+1)(n+2)=(n+1)/(n+1)+1
    Then when I tried to simplify I keep getting something that isn't an equality. Have I done something wrong so far?
    Notice that \displaystyle \begin{align*} \frac{1}{k(k + 1)} = \frac{1}{k} - \frac{1}{k + 1} \end{align*}, so your sum

    \displaystyle \begin{align*} \frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + \frac{1}{3 \cdot 4} + \dots + \frac{1}{n(n + 1)} &= \left(\frac{1}{1} - \frac{1}{2}\right) + \left(\frac{1}{2} - \frac{1}{3}\right) + \left(\frac{1}{3} - \frac{1}{4} \right) + \dots + \left(\frac{1}{n} - \frac{1}{n + 1}\right) \\ &= 1 - \frac{1}{n + 1} \\ &= \frac{n + 1}{n + 1} - \frac{1}{n + 1} \\ &= \frac{n}{n+1} \end{align*}
    Thanks from HallsofIvy
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    Re: Sigma Proof by Induction

    Quote Originally Posted by Prove It View Post
    Notice that \displaystyle \begin{align*} \frac{1}{k(k + 1)} = \frac{1}{k} - \frac{1}{k + 1} \end{align*}, so your sum

    \displaystyle \begin{align*} \frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + \frac{1}{3 \cdot 4} + \dots + \frac{1}{n(n + 1)} &= \left(\frac{1}{1} - \frac{1}{2}\right) + \left(\frac{1}{2} - \frac{1}{3}\right) + \left(\frac{1}{3} - \frac{1}{4} \right) + \dots + \left(\frac{1}{n} - \frac{1}{n + 1}\right) \\ &= 1 - \frac{1}{n + 1} \\ &= \frac{n + 1}{n + 1} - \frac{1}{n + 1} \\ &= \frac{n}{n+1} \end{align*}
    I have to prove by induction though, so I couldn't use this approach, correct?
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    Re: Sigma Proof by Induction

    Quote Originally Posted by renolovexoxo View Post
    I have to prove by induction though, so I couldn't use this approach, correct?
    Here is what is needed.
    \frac{n}{n+1}+\frac{1}{(n+1)(n+2)}=\frac{n^2+2n+1}  {(n+1)(n+2)}=\frac{n+1}{(n+2)}
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    Re: Sigma Proof by Induction

    Quote Originally Posted by renolovexoxo View Post
    I have to prove by induction though, so I couldn't use this approach, correct?
    That's not what you said. You said that you were trying to prove the statement and that you used an induction approach. I just showed you an easier approach...
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    Re: Sigma Proof by Induction

    I'm sorry, then I misstated my intentions. I need to use induction to complete this proof.
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    Re: Sigma Proof by Induction

    Quote Originally Posted by renolovexoxo View Post
    I'm sorry, then I misstated my intentions. I need to use induction to complete this proof.
    Did you follow reply #4?
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    Re: Sigma Proof by Induction

    I understood it, if that is what you're asking.
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