Let `A_n=(3/4)-(3/4)^2+(3/4)^3+….+(-1)^(n-1)(3/4)^n and B_n = 1-A_n. find the least odd nastural numebr s`n_0, so that B_ngtA_n Aangen_0`

Explanation

To solve the problem, we need to analyze the sequences \( A_n \) and \( B_n \) and find the least odd natural number \( n_0 \) such that \( B_n > A_n \) for \( n \geq n_0 \).

### Step 1: Define \( A_n \)
The sequence \( A_n \) is given by:
\[ A_n = \frac{3}{4} - \left(\frac{3}{4}\right)^2 + \left(\frac{3}{4}\right)^3 - \ldots + (-1)^{n-1} \left(\frac{3}{4}\right)^n \]
This is a finite geometric series with the first term \( a = \frac{3}{4} \) and the common ratio \( r = -\frac{3}{4} \).

### Step 2: Sum of the Geometric Series
The sum of the first \( n \) terms of a geometric series can be calculated using the formula:
\[ S_n = a \frac{1 - r^n}{1 - r} \]
Substituting the values of \( a \) and \( r \):
\[ A_n = \frac{3/4 \left(1 - (-3/4)^n\right)}{1 - (-3/4)} = \frac{3/4 \left(1 - (-3/4)^n\right)}{7/4} = \frac{3}{7} \left(1 - (-3/4)^n\right) \]

### Step 3: Define \( B_n \)
From the problem statement, we have:
\[ B_n = 1 - A_n \]
Substituting the expression for \( A_n \):
\[ B_n = 1 - \frac{3}{7} \left(1 - (-3/4)^n\right) = 1 - \frac{3}{7} + \frac{3}{7}(-3/4)^n = \frac{4}{7} + \frac{3}{7}(-3/4)^n \]

### Step 4: Set Up the Inequality
We need to find \( n_0 \) such that:
\[ B_n > A_n \]
This translates to:
\[ \frac{4}{7} + \frac{3}{7}(-3/4)^n > \frac{3}{7} \left(1 - (-3/4)^n\right) \]

### Step 5: Simplify the Inequality
Rearranging gives:
\[ \frac{4}{7} + \frac{3}{7}(-3/4)^n > \frac{3}{7} - \frac{3}{7}(-3/4)^n \]
Combine like terms:
\[ \frac{4}{7} + \frac{3}{7}(-3/4)^n + \frac{3}{7}(-3/4)^n > \frac{3}{7} \]
\[ \frac{4}{7} + \frac{6}{7}(-3/4)^n > \frac{3}{7} \]
Subtract \( \frac{3}{7} \) from both sides:
\[ \frac{1}{7} + \frac{6}{7}(-3/4)^n > 0 \]
This implies:
\[ (-3/4)^n > -\frac{1}{6} \]

### Step 6: Analyze the Inequality
Since \( (-3/4)^n \) is negative for odd \( n \), we need to find the smallest odd \( n \) such that:
\[ (-3/4)^n > -\frac{1}{6} \]
Taking \( n = 1, 3, 5, \ldots \) and checking:

- For \( n = 1 \):
\[ (-3/4)^1 = -\frac{3}{4} \quad \text{(not greater than -1/6)} \]
- For \( n = 3 \):
\[ (-3/4)^3 = -\frac{27}{64} \quad \text{(not greater than -1/6)} \]
- For \( n = 5 \):
\[ (-3/4)^5 = -\frac{243}{1024} \quad \text{(not greater than -1/6)} \]
- For \( n = 7 \):
\[ (-3/4)^7 = -\frac{2187}{16384} \quad \text{(greater than -1/6)} \]

### Conclusion
The least odd natural number \( n_0 \) such that \( B_n > A_n \) for \( n \geq n_0 \) is:
\[ \boxed{7} \]