Two formula proofs, two summation formula proofs

The first uses the method of counter-evidence.

When the sum of the two sides is less than the third side, it does not form a triangle.

The second is defined in the second one.

Two lines that do not intersect in the same plane are called parallel lines.

But later you can learn to intersect at infinity.

On the second floor: there are only two kinds of straight lines in the world, one is parallel and the other is intersecting and there are different faces.

Both of these are theorems, and both are true in Euclidean spaces.

According to the axiom: the line segment between two points is the shortest, so one side of the triangle is smaller than the sum of the other two, which is your proposition.

2. The fifth axiom of Euclidean space (i.e., ordinary three-dimensional space) is that there is and only one point parallel to a known straight line at one point outside the line. Because the point is outside the straight line, if the parallel line is compared to a point, it contradicts the axiom.

However, these are all artificial settings, and in non-Euclidean geometry the axiom of clause 2 does not hold, and parallel lines can intersect. And it's not the shortest line segment between two points.

I remember that there was an episode in the robot cat where the robot cat showed Yasuo what the shortest distance between two points on a piece of paper was--- the answer was 0, because space can be folded! Fold the paper and it's enough to coincide with the two points. But when we learn geometry, Euclidean space can't be distorted.

The first half of the sentence 1 is "the shortest line segment between two points".

2 Two lines that do not intersect are parallel lines.

The theorems can all be proved.

These are axioms

1. The straight line between two points is the shortest (this is a kilometer), so the sum of the two sides is greater than the third side.

2. Parallel lines are two straight lines that do not intersect each other, and there are only two kinds of straight lines in the world, one is parallel and the other is intersecting.

2.Suppose that line 1 and line 2 are compared to point a, and that line 3 is parallel to line 1 through point a, and point a contradicts the theorem on line 2 (on a plane, only a single line can be made parallel to a known line) so that the assumption is not true, so the parallel lines never intersect.

The first one can be with the help of a power series.

1/(1-x)=1+x+x^2+..x^n+..

Derivatives can be obtained on both sides.

1/(1-x)^2=1+2x+3x^2+..nx^(n-1)+.

Multiply x x on both sides.

x/(1-x)^2=x+2x^2+3x^3+..nx^n+..

Substituting x=1 2 yields.

The right side is the formula, and the left side is equal to 2

The second is the proportional series, which can be found by the formula of the proportional series, and of course it can also be found by power series.

The above equation = 1*(1-1 4 n) (1-1 4) n tends to , i.e., 1 4 n = 0

So the above equation = 4 3

1、sn= 1/2+2/2^2+3/2^3+4/2^4+..n-1)/2^(n-1)+n/2^n (1)

2sn=2+2/2+3/2^2+4/2^3+5/2^4+..n/2^(n-1) (2)

2)-(1), get:

sn=2+1/2+1/2^2+1/2^3+1/2^4+..1/2^(n-1)-n/2^n

From the second term to the penultimate term is a proportional series, so.

sn=2+1/2(1-1/2^(n-1))/1-1/2)-n/2^n2+1-1/2^(n-1)-n/2^n

3-1/2^(n-1)-n/2^n)

When n tends to infinity, sn=3

2、sn=1+1/4+1/4^2+..1 4 n proportional series.

1-1/4^n)/(1-1/4)

4/3（1-1/4^n）

When n tends to infinity, sn=4 3

In the past, when I was taking the Olympiad math competition, I was often discouraged and thought that I would definitely not be able to do it, so I didn't dare to imagine it boldly. Don't dare to be bold enough to study; I don't dare to be bold enough to calculate. Yes, the final result was not satisfactory either. Now, I am full of self-confidence and dare to guess everything, which is a good result for me.

It taught me that I have to have faith in everything I do, and that without faith I can't achieve anything.

A two-dimensional vector is used as an example to illustrate the process.

f(x,y)=pesai(x,y)

g(x,y)=fai(x,y)

f(x,y) +g(x,y)]'x =f'x(x,y) +g'x(x,y)

f(x,y) +g(x,y)]'y =f'y(x,y) +g'y(x,y)

Hence the gradient of f+g = the gradient of f +the gradient of g.

f(x,y) g(x,y)]'x =g(x,y)f'x(x,y) +f(x,y)g'x(x,y)

f(x,y) g(x,y)]'y =g(x,y)f'y(x,y) +f(x,y)g'y(x,y)

Get the second formula.

In fact, this type of proof focuses more on the understanding of basic concepts, as well as on the simplified form of formulas (the notation system of linear algebra can simplify cumbersome mathematical formulas).

It's kind of like the derivative of the product of two functions. I'm also crossing the line, but I haven't made an appointment there yet.

(1)a(n+1) -a(1) = nd

a(n+2) -a(2) = nd

.a(n+i) -a(i) = nd

s(2n)-s(n) = n^2 d

The same goes for the rest. 2)s(2n) = a1 + a(2) +a(n) +a(n+1) +a(n+2) +a(2n)

s(2n) = a(2n) +a(2n-1) +a(n+1) +a(n) +a(n-1) +a(1)

2s(2n) = 2n ( a1 + a(2n) )= 2n(a(n) +a(n+1))

s(2n) = n(a(n)+a(n+1))

a(2)-a(1) = d

a(4)-a(3) = d

a(2n) -a(2n-1) = d

S-even - S-odd = nd

s odd = n[a(1) + a(2n-1)].

Seven = n[a(2) +a(2n)].

s odd s even = (a(1)+a(2n-1)) (a(2)+a(2n)) = a(n) a(n+1).

s(2n-1) = s(2n) -a(2n) = n a(n) +n a(n+1) -a(2n) = 2n a(n) +nd - a(2n)

2n-1) a(n)

s odd - s even = a(2n) -s(2n) even - s(2n) odd] = a(2n) -nd = a(n).

s odd s even = (a(1)+a(2n-1)) (a(2)+a(2n-2)) = na(n) [(n-1)a(n)] = n (n-1).

Let's look at the first formula first, because both sides are positive, you can square it first, and the left side is a + 2ab + b

On the right is a +2 a b + b a must be equal to a, and b is the same.

So the left side can be seen as 2ab, the right side can be seen as 2 a b and the right side must be a non-negative number, and the left side may be equal to 2 a b, so 2ab 2 a b

So left, right.

The same is true for other formulas, all square them first and then compare the size.

a b has a b 0

a b has a b greater than 0

This line of thought to prove.

Let a+1 = x, let the left side of the equation = x 0 + x 1 + x (n-1) = y times x, then x 1+x 2+.x n = xy subtraction ==> xy-y = x n - 1==> left side of the equation = y = (x n - 1) (x-1) = ((a+1) n - 1) a = right side of the equation.

Counter-example: the side length is a right triangle with a semi-scattered diameter of 1If the bending die c is the right angle, ac-bc=1, ao-bo is not equal to 1Therefore, the original proposition is not valid.