Question 1091859
The Nth Triangular number is the sum of the sequence of positive integers up to and including N:
Noting that  A1 = 1  and An = n:

Tn = {{{ sum(k, k=1, n) = (1/2)n(A1 + An) = (1/2)n(n+1) }}} 
(you are just picking out one Tn, not the sum of an entire sequence of triangular numbers)

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That concludes the proof IF you are allowed to use the starting point  Sn=n(A1+An)/2.

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Here is more of a ground-up method that also proves Tn = n(n+1)/2 and then Sn=n(A1+An)/2:

{{{ S[1] = 1 }}}                  — also the 1st triangular number
{{{ S[2] = 1+2 = 3 }}}      — also the 2nd triangular number
: :                :
{{{ S[n] = 1+2 }}} + … + {{{+(n-1) + n }}}  — also the  n'th triangular number
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Looking at {{{ S[n] }}} you can take the largest number and pair it with the smallest (1) and form n+1, then (n-2) and 3 to get n+1, etc.  You can do this  n/2 times*.  Thus the sum 1+2+…+n  is  n(n+1)/2.
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* To see the "n/2 times" because it may not be obvious (also n may be odd) just do this:  form the pairs to be summed as (1,n), (2,n-1), (3,n-2), … ,  (n-2,3), (n-1,2), (n,1)  and note that we show each pair TWICE.

The first number in each pair counts how many pairs you have:  n
The sum of each pair is:  n+1
The sum of all the pairs shown:   n*(n+1)
Divide by 2 because we double-counted:    Sn = n*(n+1)/2   (done)
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Proof of  Sn = n(A1+An)/2:

    Sn = A1+A2+…+An

Pair them up as we did above:  (A1,An), (A2,An-1), … ,(An-1,A2), (An,A1)

Thats n pairs.
Sum of each pair is  A1+An   (A2 + An-1 = A1+An because A2 = A1 + d and An-1 = An - d,  similar logic for other pairs)
Adjust for double counting:  Sum =  n*(A1+An)/2   

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There are other ways to do the proofs, this is just the way I think about it.