$$ \sum _ { n = 0 } ^ { m i n } 3 ( 2 ) ^ { 2 p - 3 }

Question

$$ \sum _ { n = 0 } ^ { m i n } 3 ( 2 ) ^ { 2 p - 3 } < 8000 $$

UpStudy AI · Accepted Answer

Certainly! Let's break down and analyze the inequality you've provided:

$$
\sum_{n=0}^{\text{min}} 3 \cdot 2^{2p - 3} < 8000
$$

However, there seems to be some ambiguity in the expression, particularly with the upper limit of the summation, denoted as `min`. I'll proceed under the assumption that the summation is from $$ n = 0 $$ to $$ m $$, making the expression:

$$
\sum_{n=0}^{m} 3 \cdot 2^{2p - 3} < 8000
$$

If this assumption doesn't align with your intention, please provide additional clarification.

### Step 1: Understanding the Summation

The summation $$\sum_{n=0}^{m} 3 \cdot 2^{2p - 3}$$ involves adding the same term $$3 \cdot 2^{2p - 3}$$ repeatedly for each integer value of $$n$$ from $$0$$ to $$m$$. Since the term being summed does **not** depend on $$n$$, the summation simplifies to:

$$
\sum_{n=0}^{m} 3 \cdot 2^{2p - 3} = (m + 1) \cdot 3 \cdot 2^{2p - 3}
$$

### Step 2: Setting Up the Inequality

Given the simplification, the inequality becomes:

$$
(m + 1) \cdot 3 \cdot 2^{2p - 3} < 8000
$$

To make it easier to work with, let's solve for one of the variables. I'll demonstrate how to solve for $$p$$ in terms of $$m$$. If you intend to solve for a different variable, the approach will be similar.

### Step 3: Solving for $$p$$

1. **Isolate the Exponential Term:**

$$
   (m + 1) \cdot 3 \cdot 2^{2p - 3} < 8000
   $$
   
   Divide both sides by $$3(m + 1)$$:

$$
   2^{2p - 3} < \frac{8000}{3(m + 1)}
   $$

2. **Apply Logarithms:**

To solve for $$p$$, take the logarithm base 2 of both sides:

$$
   2p - 3 < \log_2\left(\frac{8000}{3(m + 1)}\right)
   $$

3. **Solve for $$p$$:**

$$
   2p < 3 + \log_2\left(\frac{8000}{3(m + 1)}\right)
   $$
   
   $$
   p < \frac{3 + \log_2\left(\frac{8000}{3(m + 1)}\right)}{2}
   $$

### Step 4: Example Calculation

Suppose you want to find $$p$$ when $$m = 10$$:

1. **Substitute $$m = 10$$:**

$$
   p < \frac{3 + \log_2\left(\frac{8000}{3(10 + 1)}\right)}{2} = \frac{3 + \log_2\left(\frac{8000}{33}\right)}{2}
   $$

2. **Calculate the Logarithm:**

$$
   \frac{8000}{33} \approx 242.424
   $$
   
   $$
   \log_2(242.424) \approx 7.91 \quad \text{(since } 2^7 = 128 \text{ and } 2^8 = 256\text{)}
   $$

3. **Compute $$p$$:**

$$
   p < \frac{3 + 7.91}{2} = \frac{10.91}{2} \approx 5.455
   $$

Since $$p$$ is typically an integer in many contexts, $$p$$ would need to be less than or equal to 5.

### Step 5: General Insights

- **Nature of $$p$$:** The variable $$p$$ is in the exponent, indicating an exponential relationship. Small changes in $$p$$ can significantly affect the sum.
  
- **Dependency on $$m$$:** As $$m$$ increases, for the inequality to hold, $$p$$ must decrease, and vice versa.

- **Constraints:** Ensure that $$m$$ is chosen such that the denominator $$3(m + 1)$$ does not make the argument of the logarithm non-positive, which would be mathematically invalid.

### Conclusion

To solve the inequality $$\sum_{n=0}^{m} 3 \cdot 2^{2p - 3} < 8000$$ for a specific variable, follow these steps:

1. Simplify the summation.
2. Isolate the exponential term.
3. Apply logarithms to solve for the desired variable.
4. Substitute known values to find specific solutions.

If you intended a different interpretation of the original summation or need to solve for a different variable, please provide further details!
To solve the inequality $$ \sum_{n=0}^{m} 3 \cdot 2^{2p - 3} < 8000 $$, follow these steps: 1. **Simplify the Summation:** $$ \sum_{n=0}^{m} 3 \cdot 2^{2p - 3} = (m + 1) \cdot 3 \cdot 2^{2p - 3} $$ 2. **Set Up the Inequality:** $$ (m + 1) \cdot 3 \cdot 2^{2p - 3} < 8000 $$ 3. **Solve for $$p$$:** $$ p < \frac{3 + \log_2\left(\frac{8000}{3(m + 1)}\right)}{2} $$ 4. **Example Calculation (when $$m = 10$$):** $$ p < \frac{3 + \log_2\left(\frac{8000}{33}\right)}{2} \approx 5.455 $$ So, $$p$$ must be less than approximately 5.455. **Conclusion:** The value of $$p$$ must be less than $$ \frac{3 + \log_2\left(\frac{8000}{3(m + 1)}\right)}{2} $$ to satisfy the inequality.

\( \sum _ { n = 0 } ^ { m i n } 3 ( 2 ) ^ { 2 p - 3 } < 8000 \)

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