Alaina's blog, 12 Sept 2010

A)

• n=1-1:1 to get next integer n=2 1+(n=2)
• n=2-2:3 to get next integer n=3 2+1+(n=3)
• n=3-3:6 to get next integer n=4 3+2+1+(n=4)
• n=4-4:10 to get next integer n=5 4+3+2+1+(n=5)
• n=5-5:15 to get next integer n=6 5+4+3+2+1+(n=6)
• n=6-6:21 to get next integer n=7 6+5+4+3+2+1+(n=7)
• n=7-7:28 to get next integer n=8 7+6+5+4+3+2+1+(n=8)
• n=8-8:36 to get next integer n=9 8+7+6+5+4+3+2+1+(n=9)
• n=9-9:45 to get next integer n=10 9+8+7+6+5+4+3+2+1+(n=10)
• n=10-10:55
  

B)

Yes, there is most definitely a pattern. I provided the work that shows the pattern the sum of integers [1,2,3,4,5,6,7,8,9,10] follows. The first integer will follow numerical order and the second will be the sum of previous integers.

C/D)

It is possible to have a formula for this type of problem. First, a reason for having a formula for this would be because NO ONE wants to sit down and add all of the integers up to 1,000,000. Since there is a common pattern to arrive at the next integer, the has to be a formula. However, some series do not follow any pattern and, therefore, do not have a formula. The formula that this particular series follows would be [n(n+1)/2] only having to plug in for "n". If you plugged in 1,000,000- [1,000,000(1,000,000+1)/2]=[1,000,000(1,000,001)]/2= 500,000,500,000.This formula proves accurate for natural and positive integers.