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Math Help - Basic proof.

  1. #1
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    Basic proof.

    Proof that the rth term can be found by finding the difference between the sum of the first r terms and the sum of the first (r-1) terms: That is, U_{r} = S_{r}-S_{r-1}.

    What I really don't understand is S_{r-1}. For example, what is the S_{r-1} of 3+5+7+9? Is it 0+3+5+7? Or 1+3+5+7?
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  2. #2
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    S_{r} is the sum of terms up to and including U_{r}
    S_{r-1} is the sum of terms up to and including U_{r-1}

    i.e.
    S_{r}=U_{1}+U_{2}+U_{3}+...+U_{r-1}+U_{r} -----(1)
    S_{r-1}=U_{1}+U_{2}+U_{3}+...+U_{r-2}+U_{r-1} -----(2)

    (1)-(2): U_{r}=S_{r}-S_{r-1}
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  3. #3
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    Quote Originally Posted by acc100jt View Post
    S_{r} is the sum of terms up to and including U_{r}
    S_{r-1} is the sum of terms up to and including U_{r-1}

    i.e.
    S_{r}=U_{1}+U_{2}+U_{3}+...+U_{r-1}+U_{r} -----(1)
    S_{r-1}=U_{1}+U_{2}+U_{3}+...+U_{r-2}+U_{r-1} -----(2)

    (1)-(2): U_{r}=S_{r}-S_{r-1}
    Yeah. I see. But how on Earth would you come up with that first? I mean, say you wanted to come up with this formula for the rth term, and you don't know that it's S_{r}-S_{r-1}? What we did now there was to confirm that U_{r} = S_{r}-S_{r-1}. There is no way I could have simply thought: Ah, find the difference between the sum of the first r terms and the sum of the first (r-1) terms.
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  4. #4
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    Look at how S_{r} is defined.

    <br />
S_{r}=U_{1}+U_{2}+U_{3}+...+U_{r-1}+U_{r}<br />

    Then consider
    <br />
S_{r-1}=U_{1}+U_{2}+U_{3}+...+U_{r-2}+U_{r-1}<br />

    If I take the difference, then I get a formula for the rth term, U_{r}=S_{r}-S_{r-1} right?
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  5. #5
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    Quote Originally Posted by acc100jt View Post
    Look at how S_{r} is defined.

    <br />
S_{r}=U_{1}+U_{2}+U_{3}+...+U_{r-1}+U_{r}<br />

    Then consider
    <br />
S_{r-1}=U_{1}+U_{2}+U_{3}+...+U_{r-2}+U_{r-1}<br />

    If I take the difference, then I get a formula for the rth term, U_{r}=S_{r}-S_{r-1} right?
    Oh, right.

    I didn't see the U_{r} at the end of the S_{r} defintion.

    Many thanks, Acc100jt.
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