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Math Help - Sum of odd integers

  1. #1
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    Sum of odd integers

    How to prove that 1+3+5+...+n = \frac{(n+1)^2}{4} ?
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  2. #2
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    In a word: induction.

    In a thousand words:
    Code:
    1 3 5 7 9
    3 3 5 7 9
    5 5 5 7 9
    7 7 7 7 9
    9 9 9 9 9
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  3. #3
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    Hello, OReilly!

    Here's one way . . .

    How to prove that: 1+3+5+...+n = \frac{(n+1)^2}{4} ?
    This is an arithmetic series
    . . with first term a = 1, common difference d = 2, and \frac{n+1}{2} terms

    Using the Sum formula: . S \:= \:\frac{1}{2}\cdot\frac{n+1}{2}\left[2\cdot1 + 2\cdot\left(\frac{n+1}{2} - 1\right)\right]

    . . . = \;\frac{n+1}{4}[2 + (n + 1 - 2)] \;= \;\frac{n+1}{4}(n+1) \;= \;\frac{(n+1)^2}{4}
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