What is the sum of the first n numbers?

Answer 1

Assuming you mean the first n counting numbers then:

let S{n} be the sum; then:

S{n} = 1 + 2 + ... + (n-1) + n

As addition is commutative, the sum can be reversed to give:

S{n} = n + (n-1) + ... + 2 + 1

Now add the two versions together (term by term), giving:

S{n} + S{n} = (1 + n) + (2 + (n-1)) + ... + ((n-1) + 2) + (n + 1)

→ 2S{n} = (n+1) + (n+1) + ... + (n+1) + (n+1)

As there were originally n terms, this is (n+1) added n times, giving:

2S{n} = n(n+1)

→ S{n} = ½n(n+1)

The sum of the first n counting numbers is ½n(n+1).

Answer 2

The sum of the first n counting numbers is n*(n+1)/2