Question 603123
{{{S=sum(k,k=50,n)}}}=50+51+52+ ... +(n-2)+(n-1)+n is a sum of n-49 terms.
It can also be written as:
{{{S=sum(k,1,n)-sum(k,1,49)}}}={{{sum(k,0,n)-sum(k,0,49)}}} 
There are many ways to approach the problem using "pre-cooked" formulas and results.
Otherwise, cooking from scratch,
{{{2*S=S+S}}}=50+51+52+ ... +(n-2)+(n-1)+n+50+51+52+ ... +(n-2)+(n-1)+n, with 2(n-49) terms.
Pairing terms (associative property) "head to tail" (first with last, second with second last, and so on), we get the sum of n-49 pair sums (n-49 sums of two numbersin brackets)
2S=(50+n)+(51+(n-1))+(52+(n-2)+ ...=50+50+50+ ... = 50(n-49)
So, if {{{2*S=50(n-49)}}} ---> {{{S=highlight(25(n-49))}}}