Heather's #2

:D YESSSSSSSSSSS! I know how to do this because Alex taught me this for my sequences and series test at nationals!! Ha anywayyyy this is what you would get using this method > (*n=6 >> 1+2+3+4+5+6)
A.)
n=1: 1
n=2: 3
n=3: 6
n=4: 10
n=5: 15
n=6: 21
n=7: 28
n=8: 36
n=9: 45
n=10: 55

B.) Yes there is definitely a pattern. First if you look at the row of sums, you notice that you get the next number by adding the next "n"...As in if you take 1, your first sum, and add 2 you get your second sum which is 3. Adding 3 to your second sum you get 6, your third sum. Then adding 4 to 6 you get your 4th sum which is 10 and so on...So in other words (looking at part A) you would add the two red numbers to get the blue number. Also, looking more closely at the sums, I notice that each n is either a divisor of or has a divisor of the sum number. For instance, for n=7, 7 is a divisor of 28..meaning that 7 can go into 28 evenly. Also, for n=4, 2 is a divisor of both 4 and 10 since 4 cannot go into 10 evenly.

C./D.) Yes it is possible to have a formula for this type of problem...but seeing as how I learned it already, I'll explain why I think there is a formula for this. First of all, I see a reason for having a formula for this because we all know that no one wants to sit down and add up all the numbers up to 1,000,000 and surely no mathematician had the time for that either. I assumed there had to be a formula since there is a common pattern here. No matter how large the sum becomes, the pattern is still there. However, that isn't always the case, but I figured there must be a pattern here since the sums are in ascending order. Sooo with that, the formula I learned is n(n+1)/2. (I honestly did learn this, I didn't look it up) To use it, all you have to do is plug in n. So if I really do want to find the sum of the first 1,000,000 natural numbers then I would get 1,000,000(1,000,001)/2 which would simplify to 500000500000. I'd say this formula is pretty accurate for natural numbers/positive integers specifically; however I'm not sure if it would work the same say if you were asked to find the sum of the first 100 negative integers..