![(a + b)^n = \sum_{k=0}^{n}\left[\binom{n}{k} a^k b ^{(n - k)} \right]](http://l.wordpress.com/latex.php?latex=%28a+%2B+b%29%5En+%3D+%5Csum_%7Bk%3D0%7D%5E%7Bn%7D%5Cleft%5B%5Cbinom%7Bn%7D%7Bk%7D+a%5Ek+b+%5E%7B%28n+-+k%29%7D+%5Cright%5D++&bg=f2e2c1&fg=000000&s=2 "(a + b)^n = \sum_{k=0}^{n}\left[\binom{n}{k} a^k b ^{(n - k)} \right]")
Proof:
When n = 1
Assume that (i) is true for n = m.
Observe that:
From (ii), (iii) & (iv)
![(a+b)^{m+1}=a^{m+1} + \sum_{k=1}^m [\binom{m}{k} + \binom{m}{k-1}] a^k b^{m + 1 - k} + b^{m+1}](http://l.wordpress.com/latex.php?latex=%28a%2Bb%29%5E%7Bm%2B1%7D%3Da%5E%7Bm%2B1%7D+%2B+%5Csum_%7Bk%3D1%7D%5Em+%5B%5Cbinom%7Bm%7D%7Bk%7D+%2B+%5Cbinom%7Bm%7D%7Bk-1%7D%5D+a%5Ek+b%5E%7Bm+%2B+1+-+k%7D+%2B+b%5E%7Bm%2B1%7D&bg=f2e2c1&fg=000000&s=0 "(a+b)^{m+1}=a^{m+1} + \sum_{k=1}^m [\binom{m}{k} + \binom{m}{k-1}] a^k b^{m + 1 - k} + b^{m+1}")
QED
Principle of finite Induction.
Let S be a set of positve integers with the properties
- 1 belongs to S, and
- whenever the integer k is in S, then the next integer k + 1 must also be in S.
Then S is the set of all positive integers.
Proof: Let T be the set of all positive integers not in S, and assume that T is nonempty. The Well-Ordering Principle tells us that T possesses a least element, which we denote by a. Since 1 is in S, certainly a > 1 so 0 < a – 1 < a. The choice of a as the smallest positive integer in T implies that a – 1 is not a member of T, or equivalently, that a – 1 belongs to S. By hypothesis, S must also contain (a – 1) + 1 = a, which contradicts the fact a lies in T. We conclude that the set T is empty, and in consequence that S contains all the positive integers.
Archimedean Property
If a and b are any positive intergers, then there exists a positive integer n such that
Proof:
Assume that the statment of the theorem is not true, so that for some a and b, 
consists entirely of positive integers. By the Well-Ordering Principle S will posses a least element, say 
contrary to the choice of
Well Ordering Principle
Every nonempty set S of nonnegative integers contains a least element; that is; there is some integer a in S such that 
Kids, kids, me, kids, kids …
Group Photo
Sweet friends.
I want to be in the photograph.
(courtesy Suman)
Had a fun day yesterday, outing to Cubbon park to manage (play with) parikrama kids. Played with kids, lots of them. They were tiny and energetic (like electrons 
). Check them out at my flickr stream. I got a marble from a kid.
12 Step NAS
I am a member of this group called Nikon-D50. NAS == Nikcon Acquisition Syndrome. Here is what a member named Scott Gibson writes about how to get NAS.
12 step NAS program.
Step 1 50mm F1.4 or f1.8
Step 2 18-200 vr
Step 3 SB 600 or SB 800
Step 4 second VR lens
Step 5 first wide angle lens
Step 6 SB 800
Step 7 second wide angle lens
Step 8 105mm Micro
step 9 Upgrade Camera Body
step 10 R1C1 lighting kit
step 11 Long prime lens
step 12 Second Micro or long prime lens
