$$1+2+3+\ldots+n=\sum_{x=1}^n x^{\underline{1}}=\frac{x(x-1)}{2}\;\Rule{1px}{4ex}{2ex}^{n+1}_{1}=\frac{(n+1)n}{2}.$$

| | | | created: 2020-05-17 11:03:05 | modified: 2020-05-17 11:09:20 | by: *bookofproofs* | references: [8404]

[8404] **Miller, Kenneth S.**: “An Introduction to the Calculus of Finite Differences And Difference Equations”, Dover Publications, Inc, 1960