Need a help in Mathematics ....?

You assume convergence. Can't do that. Suppose there are 6 terms only

x=1+2+4+8+16+32
x-1=2+4+8+16+32
x-1=2(1+2+4+8+16) .... but now the thing inside the parentheses is missing 32
x-1=2(x-32)
x-1=2x-64
x=63
In your series the "missing 32" is actually missing infinitely much.

x = 1 + 2 + 4 + 8 + 16 + 32 + …… + 2^(n-1)
x - 1 = 2 + 4 + 8 + 16 + 32 + …… + 2^(n-1)
x - 1 = 2(1 + 2 + 4 + 8 + 16 + …… + 2^n)
x - 1 ≠ 2x because x ≠1 + 2 + 4 + 8 + 16 + …… + 2^n

Here comes a mistake when you write x - 1 = 2x


====
Solution to the question :

x = 1 + 2 + 4 + 8 + 16 + 32 + …… + 2^(n-1) …… [1]

2x = 2 + 4 + 8 + 16 + 32 + 64 + …… + 2^(n-1) + 2^(n)
2x = [1 + 2 + 4 + 8 + 16 + 32 + …… + 2^(n - 1)] + [2^(n) - 1]
2x = x + [2^(n) - 1]
x = [2^(n) - 1]

The error is in your definition of x. You have not placed a limit on the summation, therefore

x=1+2+4+8+16+32+........ is actually an infinite sum.

Hence it is not true that x - 1 = 2x, both sides are STILL infinite sums, that is,

x - 1 = 2x is just saying that Infinity = infinity.

"Infinity" is not a number. You cannot do normal arithmetic with it.

what you have is an exponential equation, where the base power is 2...

2^0 = 1
2^1 = 2
2^2 = 4
2^3 = 8

and so on and so forth.

x = 2^n