What is the significance of all kinds of mathematical conjectures?

Any even number not less than 6 is the sum of two odd prime numbers; any odd number not less than 9 is the sum of three odd prime numbers. This is the famous Goldbach conjecture. Do you see that this fact is true when it is not large enough? If it is true, you have to give a reason, which leads to centuries of proof

The first four terms of 1 ^ 3,1 ^ 3 + 2 ^ 3,... Are calculated in order, from which we can guess the result of an = 1 ^ 3 + 2 ^ 3 + 3 ^ 3 +... + n ^ 3, and prove it by mathematical proof

1^2+2^2+3^2+…… +n^2=n(n+1)(2n+1)/6=[n(n+1)/2]^2
The attached certificate is as follows:
n^3-(n-1)^3=1*[n^2+(n-1)^2+n(n-1)]
=n^2+(n-1)^2+n^2-n
=2*n^2+(n-1)^2-n
2^3-1^3=2*2^2+1^2-2
3^3-2^3=2*3^2+2^2-3
4^3-3^3=2*4^2+3^2-4
.
n^3-(n-1)^3=2*n^2+(n-1)^2-n
Add all the equations
n^3-1^3=2*(2^2+3^2+...+n^2)+[1^2+2^2+...+(n-1)^2]-(2+3+4+...+n)
n^3-1=2*(1^2+2^2+3^2+...+n^2)-2+[1^2+2^2+...+(n-1)^2+n^2]-n^2-(2+3+4+...+n)
n^3-1=3*(1^2+2^2+3^2+...+n^2)-2-n^2-(1+2+3+...+n)+1
n^3-1=3(1^2+2^2+...+n^2)-1-n^2-n(n+1)/2
=3(1^2+2^2+...+n^2)=n^3+n^2+n(n+1)/2=(n/2)(2n^2+2n+n+1)
=(n/2)(n+1)(2n+1)
1^3+2^3+3^3+…… +n^3=[n(n+1)/2]^2
(n+1)^4-n^4=[(n+1)^2+n^2][(n+1)^2-n^2]
=(2n^2+2n+1)(2n+1)
=4n^3+6n^2+4n+1
2^4-1^4=4*1^3+6*1^2+4*1+1
3^4-2^4=4*2^3+6*2^2+4*2+1
4^4-3^4=4*3^3+6*3^2+4*3+1
.
(n+1)^4-n^4=4*n^3+6*n^2+4*n+1
What is the sum of all forms
(n+1)^4-1=4*(1^3+2^3+3^3...+n^3)+6*(1^2+2^2+...+n^2)+4*(1+2+3+...+n)+n
4*(1^3+2^3+3^3+...+n^3)=(n+1)^4-1+6*[n(n+1)(2n+1)/6]+4*[(1+n)n/2]+n
=[n(n+1)]^2
1^3+2^3+...+n^3=[n(n+1)/2]^2

If you're good at math, please help me figure out the winning rate
I'm 54% victorious now. How many games do I have to win to 55% without losing
22921 wins 12500 loses 9880 survives 6463

You can win 228 games in a row at most

What is the significance of all kinds of mathematical conjectures?