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Thread: sum of the integers

  1. #1
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    sum of the integers

    Prove that the sum of all positive integers between m and n inclusive (n>m) is 1/2(m+n)(n-m+1).
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  2. #2
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    Sum of AP

    Hello thereddevils
    Quote Originally Posted by thereddevils View Post
    Prove that the sum of all positive integers between m and n inclusive (n>m) is 1/2(m+n)(n-m+1).
    Do you know how to find the sum of an arithmetic progression (AP)? One formula (there are other versions of it) you can use is

    $\displaystyle S = \frac{n}{2}(a+l)$

    where $\displaystyle a$ and $\displaystyle l$ are the first and last terms, and $\displaystyle n$ is the number of terms.

    Now we want the sum of all the integers between $\displaystyle m$ and $\displaystyle n$ inclusive. So that is

    $\displaystyle m + (m+1) + ... + n$

    And this is an AP where $\displaystyle a = m, l = n$ and the number of terms is $\displaystyle n - (m-1) = n - m + 1$

    So plug these values into the formula above, and the sum is

    $\displaystyle \tfrac{1}{2}(m+n)(n-m+1)$

    Grandad
    Last edited by Grandad; Feb 23rd 2009 at 02:51 AM. Reason: Simplified solution
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