I define a distinct sum as the addition of two primes such as 3+5, 3+7, etc. The addition of two identical primes is not a distinct sum. I define presentations of an even number as the number of distinct sums this number is equal to eg. 1+19=20, 3+17=20, 7+13=20, we say the even number 20 has three presentations. I define a column as a sequence of distinct sums containing each odd prime to which is added all primes greater than this prime in a given quantity 10^n where n a non zero natural number. For example, for n=1, 1+3, 1+5, 1+7 and 3+5, 3+7, are two columns within 10^1. I define a map as the number of columns which are contained within a quantity 10^n. For example, for n=2 we have 24 odd primes and the map 10^2 contains 24 columns; the first column comprises the odd number 1 to which is added all odd primes within 10^2. Maps can be calculated vertically or horizontally. In this new approach all calculations are based in the vertical sequence of distinct sums. Let’s make a map within the quantity 10^2 so we obtain:
1+3
1+5 3+5
1+7 3+7 5+7
1+11 3+11 5+11 …
1+13 3+13 5+13 …
1+17 3+17 5+17 …
. . . …
. . . …
. . . …................... 79+83
1+89 3+89 5+89 …....79+83 83+89
1+97 3+97 5+97 …... 79+97 83+97 89+97
The columns of the map contain numbers of presentations in the following sequences:
24, 23, 22, … 3, 2, 1.
Because of this structure of sequences we can calculate the number of presentations within the quantity 10^2 with the formula (f+1)f/2 where f is the number of odd primes>1. And so we have: Number of presentations=(24+1)24/2=300 within 10^2. Because we want to exclude all presentations that are greater than 10^2, for the time being we will accept that half of the number of presentations is ≤100, so Number of presentations 10^2=300/2=150.
The number of even numbers contained in the quantity 10^2, excluding 2, is 49. So for each even number contained in 10^2 we obtain almost three presentations. This does not mean that each even number has three presentations. From the prime number theorem we have 10^n/ln10^n≈f primes so
(f+1)f/2≈10^2n/2(ln(10^n))^2
If n=2 then 10^4/2 (ln100)^2≈236. This result is a close approximation of the actual number of presentations within 10^2, which is 300. The number of even numbers contained in the quantity 10^n, excluding 2, is obtained: Even numbers=(10^n/2)-1 and so we obtain
[10^2n/2[ln(10^n)^2]/[((10^n)/2)-1]=10^2n/[2(ln(10^n)^2)(10^n-2)]
Because we have accepted only half of total presentations are equal or less than 10^n the last result is written 10^2n/[2(ln(10^n))^2(10^n-2)]. The actual number of presentations in the quantity 10^2 whose distinct sums exceed 10^2 is 104 and the actual ratio is (300-104)/49=4 which means within the quantity 10^2 we have 4 presentations for each even number. From the last relation we obtain 10^4/[2((ln100)^2)98]≈2.4 which means 2.4 presentations within 10^2 for each even number. The numerical value of the ratio 10^2n/[2(ln(10^n)^2)(10^n-2)] tends to infinity as n tends to infinity.

All even numbers greater than two can be written as 2m+2 and when m takes values from one to infinity all even numbers are presented. Also the sum of two prime numbers is written as 2k+2. If we prove k takes all values m takes then the Goldbach conjecture is proved.
When p_1=4x+1 and p_2=4x_1+1 then m=[(2x+1)^2+(2x_1+1)^2-2]/2-2x^2-2(x_1)^2
When p_1=4x+1 and p_2=4x_1+3 then m=[(2x+1)^2-(2x_1+1)^2-2]/2+(2x+2)^2-2(x_1)^2
When p_1=4x+3 and p_2=4x_1+3 then m=(2x+2)^2/2+(2x_1+2)^2/2-[(2x+1)^2+(2x_1+1)^2-2]/2.
These three cases cover all the sums of prime numbers. That means k takes as many values as the number of presentations in a given quantity 10^n.
So (k number of values)/(m number of values) tends to infinity as n tends to infinity.
From the above we can show by shear calculation m takes all values k takes from 1 to 100 and from 100 to 1,000,000 and from 10^6 to 10^9 and because the relation of k to m is asymptotic, m will take all values k takes.
My question is, is this form of mathematical induction correct or incorrect? If incorrect, what are the logical arguments to support such an opinion?
If this question is not acceptable as it is, please suggest improvements.

I think it is ok know to post my question on MO.

Thanks for the suggestions. I made the chances.

Dear Scott.
I do not want you to apply mathematical induction. What I am asking is whether, from the facts presented, mathematical induction is applicable. If it is true for the first 100 even numbers and then to the first million or billion it should be true for any zillion even numbers. The facts are proving the relation of K and M is asymptotic. I hope this alleviates your concerns about the intent of my question.
"What sort of answer are you expecting?" I want to know if there is any flaw in my way of thinking.
Vassili

Dear mariano.
I edited the question and i will appreciate any comments .

Dear Scott.
Thank you for the information and your opinion. I want to bring to your attention to the following. We can measure horizontally the even numbers on each map. If we do that in the following way we always obtain a number of even numbers which do not appear again. If p_1 is the immediate prime smaller than (10^2)/2 and if p_2 is the immediate prime less than 10^2 then all even numbers equal to or smaller than (10^2)/2 do not appear after p_2.This process is an exhaustive process containing in each step a greater number of even numbers which exist within a bound set by the immediate prime numbers as above. Because of the gravity of the problem I think it is fair to be posted so others will have the opportunity to express an opinion.

Dear Mariano you and Scott you are entitled to your opinions without presenting any mathematical arguments. Every time I try to make my arguments as clear as possible, backing them with logical facts. If anyone brings a sound mathematical argument which contradicts my ideas I will accept it. If you want to force your opinions, I have no way of stopping you.

I agree with Yemon's last paragraph, and I can safely say that any question on possible proofs to Goldbach using the techniques outlined will be closed, and quickly. This is not an opinion, but borne out by experience with a) mathematics b) mathematicians and c) the rules of this site.

Dear Yemon,
Scott has asked me to present an exhaustive process which i did as you can see above. With this process we obtain the asymptotic relation as presented above. Since I received three threats you can go ahead and close the thread since you consider yourselves the only experts.

I presented several arguments M takes all values K takes. I accept is not a complete proof.I cannot think of a single argument to support the opposite. I am asking whether someone else can. Now you can close the thread despite the fact the material has a lot of substance which will be useful to some one .