\sum_(n=1)^(\infty ) ((4^(n)-3^(n))/(12^(n)))^(2)

The given infinite sum is:

\[
S = \sum_{n=1}^{\infty} \left( \frac{4^n - 3^n}{12^n} \right)^2
\]

### Step-by-step Solution:

1. **Simplify the expression inside the sum:**
   \[
   \frac{4^n - 3^n}{12^n} = \left( \frac{4}{12} \right)^n - \left( \frac{3}{12} \right)^n = \left( \frac{1}{3} \right)^n - \left( \frac{1}{4} \right)^n
   \]

2. **Square the simplified expression:**
   \[
   \left( \frac{4^n - 3^n}{12^n} \right)^2 = \left( \left( \frac{1}{3} \right)^n - \left( \frac{1}{4} \right)^n \right)^2
   \]
   Expand the square:
   \[
   \left( \left( \frac{1}{3} \right)^n - \left( \frac{1}{4} \right)^n \right)^2 = \left( \frac{1}{3^n} \right)^2 - 2 \left( \frac{1}{3^n} \cdot \frac{1}{4^n} \right) + \left( \frac{1}{4^n} \right)^2
   \]
   Simplifying each term:
   \[
   = \frac{1}{3^{2n}} - 2 \cdot \frac{1}{12^n} + \frac{1}{4^{2n}}
   \]

3. **Write the sum with the expanded terms:**
   \[
   S = \sum_{n=1}^{\infty} \left( \frac{1}{3^{2n}} - 2 \cdot \frac{1}{12^n} + \frac{1}{4^{2n}} \right)
   \]
   This can be split into three separate sums:
   \[
   S = \sum_{n=1}^{\infty} \frac{1}{3^{2n}} - 2 \sum_{n=1}^{\infty} \frac{1}{12^n} + \sum_{n=1}^{\infty} \frac{1}{4^{2n}}
   \]

4. **Use the formula for a geometric series:**
   The sum of a geometric series \(\sum_{n=1}^{\infty} ar^n = \frac{ar}{1 - r}\) for \(|r| < 1\).

   - First sum: \(\sum_{n=1}^{\infty} \frac{1}{3^{2n}}\)
     \[
     = \frac{\frac{1}{9}}{1 - \frac{1}{9}} = \frac{1}{8}
     \]
   - Second sum: \(\sum_{n=1}^{\infty} \frac{1}{12^n}\)
     \[
     = \frac{\frac{1}{12}}{1 - \frac{1}{12}} = \frac{1}{11}
     \]
   - Third sum: \(\sum_{n=1}^{\infty} \frac{1}{4^{2n}}\)
     \[
     = \frac{\frac{1}{16}}{1 - \frac{1}{16}} = \frac{1}{15}
     \]

5. **Substitute the values back into the equation for \(S\):**
   \[
   S = \frac{1}{8} - 2 \cdot \frac{1}{11} + \frac{1}{15}
   \]

6. **Combine the terms:**
   \[
   S = \frac{1}{8} - \frac{2}{11} + \frac{1}{15}
   \]
   To combine these fractions, find a common denominator, which is 1320:
   \[
   S = \frac{165}{1320} - \frac{240}{1320} + \frac{88}{1320}
   \]
   Simplifying:
   \[
   S = \frac{165 - 240 + 88}{1320} = \frac{13}{1320}
   \]
   Simplifying further:
   \[
   S = \frac{1}{101.54}
   \]
Therefore, the sum converges to approximately \( \frac{13}{1320} \).

Would you like further details on any part of this solution?

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Here are some related questions to explore:

1. How does the geometric series formula apply to infinite sums?
2. What are the conditions for an infinite series to converge?
3. How do different types of series (arithmetic, geometric, harmonic) differ?
4. What is the radius of convergence for series like this one?
5. How can you approximate the sum of a series numerically?

**Tip:** Recognizing geometric series helps simplify complex sums quickly!

Evaluate the sum: \sum_{n=1}^{\infty} \left( \frac{4^n - 3^n}{12^n} \right)^2

This solution provides a step-by-step approach to solving the infinite sum \sum_{n=1}^{\infty} \left( \frac{4^n - 3^n}{12^n} \right)^2. Key concepts include geometric series, algebraic simplification, and series convergence. Suitable for undergraduate students studying calculus or algebra.

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