If the number of terms in the expansion of `(1-2/x+4/(x^2))^n , x!=0,` is 28, then the sum of the coefficients of all the terms in this expansion, is : (1) 64 (2) 2187 (3) 243 (4) 729

Options

A. 2187
B. 243
C. 729
D. 64

Explanation

To solve the problem step by step, we need to determine the number of terms in the expansion of \((1 - \frac{2}{x} + \frac{4}{x^2})^n\) and then find the sum of the coefficients of all the terms in this expansion.

### Step 1: Determine the number of terms in the expansion
The number of terms in the expansion of \((a + b + c)^n\) is given by the formula:
\[ \text{Number of terms} = \frac{(n + r - 1)!}{(n!)(r - 1)!} \]
where \(r\) is the number of different variables (terms). In our case, we have three terms: \(1\), \(-\frac{2}{x}\), and \(\frac{4}{x^2}\). Thus, \(r = 3\).

The formula simplifies to:
\[ \text{Number of terms} = n + 1 \text{ (since } r = 3 \text{)} \]
Given that the number of terms is 28, we have:
\[ n + 1 = 28 \]
This leads to:
\[ n = 28 - 1 = 27 \]

### Step 2: Set up the equation for the number of terms
The correct formula for the number of terms in the expansion is:
\[ \frac{(n + 2)(n + 1)}{2} = 28 \]
This can be rearranged to:
\[ (n + 2)(n + 1) = 56 \]
Expanding this gives:
\[ n^2 + 3n + 2 = 56 \]
Rearranging leads to:
\[ n^2 + 3n - 54 = 0 \]

### Step 3: Solve the quadratic equation
Now we solve the quadratic equation \(n^2 + 3n - 54 = 0\) using the factorization method:
\[ n^2 + 9n - 6n - 54 = 0 \]
Grouping the terms:
\[ n(n + 9) - 6(n + 9) = 0 \]
Factoring out \((n + 9)\):
\[ (n + 9)(n - 6) = 0 \]
This gives us two solutions:
\[ n + 9 = 0 \Rightarrow n = -9 \quad \text{(not valid)} \]
\[ n - 6 = 0 \Rightarrow n = 6 \]

### Step 4: Find the sum of the coefficients
Now that we have \(n = 6\), we need to find the sum of the coefficients in the expansion of \((1 - \frac{2}{x} + \frac{4}{x^2})^6\). The sum of the coefficients can be found by substituting \(x = 1\):
\[ (1 - 2 + 4)^6 = (3)^6 \]
Calculating \(3^6\):
\[ 3^6 = 729 \]

### Conclusion
The sum of the coefficients of all the terms in the expansion is:
\[ \boxed{729} \]