Question 15836
This was a challenge question in my pre-algebra class.
(1,1) (2,3) (3,6) (4,10) (5,15).... 
 key to such problems is to analyse the pattern...each is an ordered pair 
   1st numbers in the ordered pairs are 1,2,3,4,5..etc...so each increases by 1 so graph these seperately if you want .you will find a straight line ..here     n th. number is n if we call these as x1,x2,x3,x4,x5...etc then xn=n ...( you can check by putting n=1,2,3 etc when you will get the first numbers in the pairs as 1,2,3,4,5..etc.. ) 
now 2nd nos. are 1,3,6,10,15..no constant difference or ratio is visible so try the differences between pairs of consecutive numbers..
  3-1=2
  6-3 =3
10-6=4
15-10=5...now this is a pattern ..if we call the                                1 = y1...3=y2...6=y3...10=y4...15=y5..then 
  y2-y1=2
y3-y2=3
y4-y3=4
y5-y4=5
....
....
yn-y(n-1)= n...see the pattern ..check if you want as explained above 
 
add all the above ...we get 
 yn-y1=2+3+4+5+...n 
 yn = y1+2+3+4+5+......n =1+2+3+4+5+.....n =you will learn it as equal to          n(n+1)/2 
 hence yn =n(n+1)/2 ....check if you want as explained above 
 that is put n=1,2,3,4,5...etc in n(n+1)/2.. ...we get..y1,y2,y3,y4,y5...etc.as 1,3,6,10,15...respectively