Lesson 5

The sum \(1+2+3+\dotsb+n\) is known as an arithmetic series. The formula for it is well-known.

Theorem (arithmetic series formula). For any positive integer \(n\), \[1+2+\dotsb+n=\frac{n(n+1)}{2}.\]

Proof. We prove the formula using the method of mathematical induction.

• Base case \(n=1\).  \(1=1\).


• Inductive step \(n \to n+1\).  \[\begin{align*}1+2+\dotsb+n+(n+1)&=\left(1+2+\dotsb+n\right)+(n+1)&\text{(regroup)}\\&=\frac{n(n+1)}{2}+(n+1)&\text{(inductive hypothesis)}\\&=\frac{n(n+1)}{2}+\frac{2(n+1)}{2}\\&=\frac{(n+1)(n+2)}{2}\ .\end{align*}\]

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Again, we have proven the formula without obtaining any information about where the formula itself comes from. The following two techniques will help to understand the origin of this formula.

• Arithmetic.  In the sum \(1+2+3+\dotsb+(n-2)+(n-1)+n\), it is instinctive to group the terms as follows: the first with the last, the second with the second last, and so on. In each group, the sum is \((n+1)\) and there are about \(n/2\) groups. Visually, this can be remembered as follows: write the sum once in the usual order, and below it in the reverse order. This way, you get \(n\) columns, each of which sums to \((n+1)\).
  

  



• Geometric.  If each number in the sum is associated with a vertical and horizontal strip of corresponding length, then all these strips can be combined to form a rectangle of size \((n+1) \times n\). Or this way: from strips of width from \(1\) to \(n\), you can form a figure whose area will be equal to half the sum of the area of a square with side \(n\) and the areas of \(n\) squares with side \(1\) (total \(\frac{n^2}{2}+n\cdot\frac{1}{2}\)).
  

  



• Oracular.  Of course, one may also consult Internet or LLMs.