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