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.
Can someone explain to me how this happens?
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
Adding the forward and reverse sums we get:
{: (2S=,n,+n,+n,+, ...,+n,+n ) :}
As the RHS contains
2S = overbrace(n+n+...+n)^"n-1"
\ \ \ \ = n(n-1)
Leading to the given result:
S = 1/2n(n-1) \ \ \ \ QED