Fermat’s little theorem: Let p be a prime. Then a^p-a is a multiple of p. In other words a^p is congruent to a mod p.
Proof: We will use Induction. For a=1 we have a^p-a=1^p-1=1-1=0. 0 is a multiple of p.
Now we assume, that the theorem works for a.
Last but not least, we have to show that a+1 works. We have (a+1)^p-(a+1)=(a+1)^p-a-1. Now we use the binomial theorem:
We know that a^p-a is a multiple of p. Every summand of the sum is a multiple of p as well. It follows, that the theorem works for a+1 as well.
This proves the theorem.
A few months ago, I stumbled a pond a new problem solving idea. It is called the telescope method. I did know about it but I just used it subconsciously. I want to present a rather easy problem that can be solved with this method. In 1672
Huygens asked Leibnitz to solve this problem, which he did using the same method.
The problem is this: What is the sum off all
reciprocals off the triangle numbers.
First we find a formula to express a triangle number. it is well known that this formula is: (n*(n+1))/2 . Let x be the solution to the problem. Let dn denote the nth triangle number.
Now there is a nice trick that you can use:
This has a great benefit for infinite sums. We add this into the equation above and get:
Lets se what happens to the first few n. 1/1-1/2+1/2-1/3+1/3-14+1/4-1/5. As you can see, everything except the 1/1 cancels itself out. 1/1+(-1/2+1/2)+(-1/3+1/3)+(-14+1/4)+(-1/5… =1/1+0+0+0+0….
We have 1=1/2x. Simple algebra gives us x=2, which is the solution to the problem.
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Uncategorized and tagged algebra, equation, infinity, integer, mutiplikation, number, problem, science, sum, telescope on . August 7, 2014