proof 1


  1 + 2 +     3     + ... + n-1 + n    = S   # forwards
  n + (n-1) + (n-2) + ... + 2   + 1    = S   # backwards    add
  ----------------------------------------
  (1+n) + (1+n) +  ... + (1+n)         = 2*S

  n terms

  n * (1+n)  = 2 * S


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

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proof by induction

step 1: start someplace.

True for n == 1?      n*(n+1)/2   ... 1*(1+1)/2 = 1
YES.

step 2:

assume true for n
1 + 2 + 3 + .... n =  n * (n+1) /2

show its true for n+1

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

But (n+1)*(n+2)/2 = 1/2 (  n*n + n + 2n + 2)
                   =    n*n/2 + 3/2 n + 2/2                                   
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