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    I've been struggling with the following question and have no idea how to prove it, a full explanation would be helpful:

    Prove that the sum of all the integers between m and n inclusive (n>m) is 0.5(m+n)(n-m+1).
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    (Original post by Alb1)
    I've been struggling with the following question and have no idea how to prove it, a full explanation would be helpful:

    Prove that the sum of all the integers between m and n inclusive (n>m) is 0.5(m+n)(n-m+1).
    You can prove it directly.

    \displaystyle \sum_{r=1}^{n} r = 0.5n(n+1)

    And \displaystyle \sum_{r=m}^n r = \sum_{r=1}^n r - \sum_{r=1}^{m-1} r


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    Thanks.
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    I also found a different solution on this website https://undergroundmathematics.org/s...r7424/solution
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    (Original post by Alb1)
    I also found a different solution on this website https://undergroundmathematics.org/s...r7424/solution
    That's the exact same solution to your question just using the sigma notation instead.

    Here's the link:

    \displaystyle \sum_{r=1}^n r = 1+2+3+...+(n-1)+n =\frac{1}{2}n(n+1) which is what they use and even split it for the m in the same way that I've shown.
 
 
 
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