To solve this problem, let's analyze it step by step:

1. Understand the Problem

The expansion of $(1 + x)^{12}$ follows the binomial theorem:

$(1 + x)^{12} = \sum_{k=0}^{12} \binom{12}{k} x^k$

The terms equidistant from the middle in this expansion are $T_r = \binom{12}{r} x^r$ and $T_{12-r} = \binom{12}{12-r} x^{12-r}$.

The ratio of these terms is given as:

$\frac{T_r}{T_{12-r}} = \frac{\binom{12}{r} x^r}{\binom{12}{12-r} x^{12-r}}$

Simplify the ratio:

Since $\binom{12}{r} = \binom{12}{12-r}$, the ratio becomes:

$\frac{T_r}{T_{12-r}} = \frac{x^r}{x^{12-r}} = x^{2r - 12}$

We are given that this ratio equals $\frac{1}{256}$, so:

$x^{2r - 12} = \frac{1}{256}$

Express $\frac{1}{256}$ as a power of 2:

$\frac{1}{256} = 2^{-8}$

Thus:

$x^{2r - 12} = 2^{-8}$

Take logarithms to solve for $x$:

$2r - 12 = -8 \implies 2r = 4 \implies r = 2$

2. Sum of All Terms

The sum of all terms in the binomial expansion $(1 + x)^{12}$ is obtained by substituting $x = 1$ into the expansion:

$\text{Sum} = (1 + 1)^{12} = 2^{12} = 4096$

Final Answer:

The sum of all the terms in the expansion is 4096.

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Would you like more details or have any questions?

Here are 5 related questions for further exploration:

1. How do you find the $r$-th term in a binomial expansion?
2. What is the middle term of $(1 + x)^{12}$?
3. How does the ratio of equidistant terms change if $x \neq 1$?
4. What is the general formula for the sum of all terms in $(1 + x)^n$?
5. Can you derive properties of coefficients using equidistant terms in the binomial expansion?

Tip: For ratios involving equidistant terms, consider the symmetry of binomial coefficients!