Landlord, you can use the hypothesis method to solve it (because the following equation holds for any real number x)

Solution:

(1) Let x=0.

Then there are: (1+0)+(1+0) 2+(1+0) 3+...+(1+0) n = A0+a1(0).

Namely:1+1+1+1...+1(n1) = A0+0.

Get n = a0.

(2) Let x= 1.

There are: (1+1)+(1+1) 2+(1+1) 3+(1+).

Namely: 2+2 2+2 3+...+2 n = A0+a1+A2+A3+...+An.

=2+2^2+2^3+………+2^n=+57

=2+2^2+2^3+………+2^n-57=

(2+2 2+2 3+...+2 n should have a formula, but I don't know, I can only use the assumed value of n to find it. Please forgive me! )

(3) Let n=6.

Then: 2+22+23+24+25+26-57 = n.

=2+4+8+ 16+32+64-57= n

= 126-57 = n n=69 and n=6 are contradictory.

∴n>6

Let n=5.

Then it is: 2+22+263+24+25-57 = n.

=2+4+8+ 16+32-57= n

= 62-57 = n n = 5 ∴ n = 5 hold.

∴n=5