How does 1+2+3+...+(n-1)=1/2(n)(n-1)?

Can someone explain to me how this happens?
1+2+3+...+(n-1)=1/2(n)(n-1)

Thank you.

1 Answer
Jun 5, 2018

We seek to prove that:

1+2+3+...+(n-1)=1/2(n)(n-1)

We can write the sum, and also reverse the terms:

S = 1 + 2 + 3 + ... + (n-2) + (n-1)

Writing the sum as, and in addition reversing the terms:

{: (S=,1,+2,+3,+, ...,+(n-2),+(n-1) ), (S=,(n-1),+(n-2),+(n-3),+, ...,+2,+1 ) :}

When each sum has n-1 terms.

Adding the forward and reverse sums we get:

{: (2S=,n,+n,+n,+, ...,+n,+n ) :}

As the RHS contains n-1 terms, we have:

2S = overbrace(n+n+...+n)^"n-1"

\ \ \ \ = n(n-1)

Leading to the given result:

S = 1/2n(n-1) \ \ \ \ QED