Answer
The numbers from $1\ to\ n$, where $n$ is $even$, can be grouped into $n/2$ pairs of the form
$1 + n = n + 1$
$2 + (n - 1) = n + 1$
$… $
$n /2 + (n/2 + 1) = n + 1 $
giving a sum of $(n/2)(n + 1)$.
This formula gives the correct sum for all cases shown, whether $n$ is $even$ or $odd$.
Work Step by Step
The numbers from $1\ to\ n$, where $n$ is $even$, can be grouped into $n/2$ pairs of the form
$1 + n = n + 1$
$2 + (n - 1) = n + 1$
$… $
$n /2 + (n/2 + 1) = n + 1 $
giving a sum of $(n/2)(n + 1)$.
This formula gives the correct sum for all cases shown, whether $n$ is $even$ or $odd$.
