Response 418

Gauss’s Formula for Summing Numbers Continued

In an earlier post I mentioned coming across the reasoning followed by Gauss to prove that the sum of the first n numbers is equal to n(n+1)/2. I started wondering if there is no generalization available for this formula: what is the sum of the whole number m to n, m and n included. Following the same reasoning as Gauss you can show that in this case the sum turns out to be (n+m)(n-m+1)/2. If you plug in m-1 you get the same formula as before.

Σ1n=(n+1)n2\Sigma_1^n = \frac{(n+1)n}{2}
Σmn=Σ1n−Σ1m−1=(n+1)n2−(m−1+1)(m−1)2=(n+1)n2−m(m−1)2\Sigma_m^n = \Sigma_1^n – \Sigma_1^{m-1} = \frac{(n+1)n}{2} – \frac{(m-1+1)(m-1)}{2} = \frac{(n+1)n}{2} – \frac{m(m-1)}{2}
=n2+n−m2+m2=(n2−m2)+(n+m)2=(n+m)(n−m)+(n+m)2= \frac{n^2+n-m^2+m}{2} = \frac{(n^2-m^2)+(n+m)}{2} = \frac{(n+m)(n-m)+(n+m)}{2}
=(n+m)(n−m+1)2:QED=\frac{(n+m)(n-m+1)}{2} :QED

PeeCleemens

,
, ,

Leave a comment