Post 3

So, I just realized that Jordan's the one doing this blog...I'm must have been out of it when that memo was given out. Anyway, let's get on with the math, shall we? Kay, we shall.

Question 1:
a.) What is the sum of the first n positive, non-zero, even integers?

(1) 2= 2
(2) 2+4= 6
(3) 2+4+6= 12
(4) 2+4+6+8= 20
(5) 2+4+6+8+10= 30
(6) 2+4+6+8+10+12= 42
(7) 2+4+6+8+10+12+14= 56
(8) 2+4+6+8+10+12+14+16= 72
(9) 2+4+6+8+10+12+14+16+18= 90
(10) 2+4+6+8+10+12+14+16+18+20= 110
b.) How does this compare to the sum of the first n integers (from last week?) Here I mean how does it compare numerically, e.g. When n=1, how does this some compare to last week's sum? When n=2, how do they compare? etc. By doing this, one should be able to figure out the generic formula using last week's solution.

Okay. So last week, I did point out that in the case of n=3, the sum was n^2 - n. However....in this case it's n^2 + n, when looking at the above sums, this becomes evident to me...now of course.

c.) Why does this make any logical sense? Be careful before jumping to immediate conclusions!

Well, most of the time I lack in the logic department. But I'll give it a whirl. Ummm...this makes "logical sense" because...okay. I'm at a loss.

Question 2:
Let T be sum of the first n positive, non-zero, odd integers?
Similarly to question one, this just means if n=2, T=1+3=4 and so on.
a.) Write out the sums when n=1, 2, 3, 4, 5, 6, 7, and 8?.

Sums? I like these problems..they start out easy then just progress with the whole thinking thing. gahh!

(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

Got it? good.

b.) What is the obvious pattern that arises? What is the sum for any generic number of terms, n? I am expected a formula for this part! This one is not complicated at all assuming your partial sums are correct. If you're staring at your sums and don't know the answer to this one, you most likely calculated them incorrectly!

n^2. does it really need an explanation??

c.) Justify that this holds true for all n. This step IS required and is the main question in this week's blog post. An answer of "I know it's true but don't know why" is NOT acceptable for this one!
Hint: Think about either a geometric argument or an algebraic one using last week's question.

This n^2 thing does hold true for all numbers, because if you think of it "logically", because T(n-1) + n+n-1 ..which basically says T(n-1) + 2n - 1. So THIS basically says, that the previous sum plus the 2n - 1 giving you the next odd number to just add to the previous. I think I'm being kind of redundant, but to be honest, I'm at a loss for words...