Find the sum of the first n positive integers.

The correct answer is Option (3) → \(\frac{n(n+1)}{2}\)

The sum \(S=1+2+3+\dots+n\) is found by writing it forwards and backwards and adding the two lines term by term: every pair then has the same sum.

• The sum written forwards: \(S=1+2+\dots+(n-1)+n\)

• The same sum written backwards: \(S=n+(n-1)+\dots+2+1\)

• Adding the two lines: each of the \(n\) pairs (1 and \(n\), 2 and \(n-1\), and so on) adds up to \(n+1\), so \(2S=n(n+1)\)

• So the sum is \(\displaystyle S=\frac{n(n+1)}{2}\)