Alaina's blog, 19 Sept 2010

Question 1:

A) n=1: 2=2
n=2: 2+4=6
n=3: 2+4+6=12
n=4: 2+4+6+8=20
n=5: 2+4+6+8+10=30
n=6: 2+4+6+8+10+12=42
n=7: 2+4+6+8+10+12+14=56
n=8: 2+4+6+8+10+12+14+16=72
n=9: 2+4+6+8+10+12+14+16+18=90
n=10: 2+4+6+8+10+12+14+16+18+20=110

B) looking back at last week's sums they were, starting at n=1,:
1, 3, 6, 10, 15.. With those numbers, to get the next sum number, all you have to do is add the next "n". meaning that to find the sum number after 6, where n=3, add the next "n" which would be 4. And that would give me the correct sum, 10. for this week, since the numbers are only even numbers, there's a bigger jump from one sum number to the next. First looking at only the sum #'s, starting with 2 you add 4, then you add 4, then 6, then 8, etc. to get the next sum number. you must have to multiply something together in order to get the formula. The formula I came up with to find the sums is 2n. it just sort of gets you where you need to be. Using the formula I had last week, I made a minor adjustment for it to work this time. So my formula for this is 2n(n+1)/2. all you have to do to find the sum is plug in n.

C) I'd say it makes sense because it's the same concept as what we did last week, just with a different category of numbers..So you would have to tweek the formula a little bit in order for it to work properly.

Question 2:

A) n=1: 1=1
n=2: 1+3=4
n=3: 1+3+5=9
n=4: 1+3+5+7=16
n=5: 1+3+5+7+9=25
n=6: 1+3+5+7+9+11=36
n=7: 1+3+5+7+9+11+13=49
n=8: 1+3+5+7+9+11+13+15=64

B) looking at the sums, the formula for this would be n^2 because all the sums are perfect squares.

C) This is true for all n because no matter what n is, you're going to multiply it by itself to get your sum every time. So that will not change anything. And there are really no specifics when it comes to the "category" of the number. for example if you are asked to find the sum when n=137, all you do is multiply 137 by itself and you get 18769.