Let `A_(n) = (3)/(4) - ((3)/(4))^(2) + ((3)/(4))^(3)-...+ (-1)^(n-1) ((3)/(4))^(n) and B_(n) = 1 - A_(n)`, then find the least value of `n_(0), n_(0) in N` such that `B_(n) gt A_(n), AA n ge n_(0)`.

Answer

To solve the problem, we need to analyze the sequences \( A_n \) and \( B_n \) given by:

\[ A_n = \frac{3}{4} - \left(\frac{3}{4}\right)^2 + \left(\frac{3}{4}\right)^3 - \ldots + (-1)^{n-1} \left(\frac{3}{4}\right)^n \]

\[ B_n = 1 - A_n \]

We want to find the least value of \( n_0 \) such that \( B_n > A_n \) for all \( n \geq n_0 \).

### Step 1: Find the sum \( A_n \)

The series \( A_n \) is a finite geometric series with the first term \( a = \frac{3}{4} \) and common ratio \( r = -\frac{3}{4} \).

The formula for the sum of the first \( n \) terms of a geometric series is:

\[ S_n = a \frac{1 - r^n}{1 - r} \]

Substituting the values:

\[ A_n = \frac{3}{4} \cdot \frac{1 - \left(-\frac{3}{4}\right)^n}{1 - \left(-\frac{3}{4}\right)} \]

Calculating the denominator:

\[ 1 - \left(-\frac{3}{4}\right) = 1 + \frac{3}{4} = \frac{7}{4} \]

Thus, we have:

\[ A_n = \frac{3}{4} \cdot \frac{1 - \left(-\frac{3}{4}\right)^n}{\frac{7}{4}} = \frac{3}{7} \left(1 - \left(-\frac{3}{4}\right)^n\right) \]

### Step 2: Find \( B_n \)

Now, substituting \( A_n \) into \( B_n \):

\[ B_n = 1 - A_n = 1 - \frac{3}{7} \left(1 - \left(-\frac{3}{4}\right)^n\right) \]

Simplifying \( B_n \):

\[ B_n = 1 - \frac{3}{7} + \frac{3}{7} \left(-\frac{3}{4}\right)^n = \frac{4}{7} + \frac{3}{7} \left(-\frac{3}{4}\right)^n \]

### Step 3: Set up the inequality \( B_n > A_n \)

We need to solve the inequality:

\[ \frac{4}{7} + \frac{3}{7} \left(-\frac{3}{4}\right)^n > \frac{3}{7} \left(1 - \left(-\frac{3}{4}\right)^n\right) \]

This simplifies to:

\[ \frac{4}{7} + \frac{3}{7} \left(-\frac{3}{4}\right)^n > \frac{3}{7} - \frac{3}{7} \left(-\frac{3}{4}\right)^n \]

Combining like terms gives:

\[ \frac{4}{7} + \frac{3}{7} \left(-\frac{3}{4}\right)^n + \frac{3}{7} \left(-\frac{3}{4}\right)^n > \frac{3}{7} \]

This leads to:

\[ \frac{4}{7} + \frac{6}{7} \left(-\frac{3}{4}\right)^n > \frac{3}{7} \]

Subtracting \( \frac{3}{7} \) from both sides:

\[ \frac{1}{7} + \frac{6}{7} \left(-\frac{3}{4}\right)^n > 0 \]

### Step 4: Solve for \( n \)

Now we can isolate \( \left(-\frac{3}{4}\right)^n \):

\[ \frac{6}{7} \left(-\frac{3}{4}\right)^n > -\frac{1}{7} \]

Multiplying both sides by \( \frac{7}{6} \):

\[ \left(-\frac{3}{4}\right)^n > -\frac{1}{6} \]

Since \( \left(-\frac{3}{4}\right)^n \) alternates in sign, we need to consider when \( n \) is even or odd. We find that for large \( n \), \( \left(-\frac{3}{4}\right)^n \) approaches 0.

### Step 5: Find the least \( n_0 \)

To find the least \( n_0 \), we can test small values of \( n \):

- For \( n = 1 \): \( B_1 = \frac{4}{7} + \frac{3}{7}(-\frac{3}{4})^1 = \frac{4}{7} - \frac{9}{28} = \frac{4}{7} - \frac{9}{28} = \frac{7}{28} = \frac{1}{4} < A_1 \)
- For \( n = 2 \): \( B_2 = \frac{4}{7} + \frac{3}{7}(-\frac{3}{4})^2 = \frac{4}{7} + \frac{27}{112} > A_2 \)
- Continuing this way, we find that \( n_0 = 7 \) is the least value satisfying the condition.

Thus, the least value of \( n_0 \) such that \( B_n > A_n \) for all \( n \geq n_0 \) is:

\[ \boxed{7} \]