Post 3

1)
a) What is the sum of the first n positive, non-zero, even integers?
S = (n)(n+1)

b) The sum if only doing even integers will be double than if doing all integers.

c) It makes sense because of the fact that every number you're adding for even integers is exactly double than for all integers.
i.e.
n=6
all: S = 1+2+3+4+5+6 = 21
even: S = 2+4+6+8+10+12 = 42
1*2 = 2, 2*2 = 4, 3*2 = 6, etc.

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

b)the answer will be S = n^2

c) Another pattern I found that may also be the proof is:
in all n's the first and last number added in the series combine to give you 2n. And then combined with n/2 also gives you n^2.