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

Question

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

Thank you.

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Answer 1 · 2018-06-05T20:56:17

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 #