\(\frac{15}{11}\)
The problem requires the evaluation of a summation series:
\[ 1 + \left(1 + \frac{1}{3}\right)\frac{1}{4} + \left(1 + \frac{1}{3} + \frac{1}{9}\right)\frac{1}{16} + \left(1 + \frac{1}{3} + \frac{1}{9} + \frac{1}{27}\right)\frac{1}{64} + \cdots \]
The expression within the parentheses constitutes a geometric series with first term \( a = 1 \) and common ratio \( r = \frac{1}{3} \), comprising \( n \) terms. The sum of a finite geometric series is given by:
\[ S_n = \frac{a(1 - r^n)}{1 - r} \Rightarrow \frac{1 - \left(\frac{1}{3}\right)^n}{1 - \frac{1}{3}} = \frac{1 - \left(\frac{1}{3}\right)^n}{\frac{2}{3}} = \frac{3}{2} \left(1 - \left(\frac{1}{3}\right)^n\right) \]
Each of these sums is multiplied by \( \frac{1}{4^n} \), resulting in the complete series:
\[ 1 + \sum_{n=1}^{\infty} \left( \frac{3}{2} \left(1 - \left(\frac{1}{3}\right)^n \right) \cdot \frac{1}{4^n} \right) \]
The summation is decomposed into two parts:
\[ 1 + \frac{3}{2} \left( \sum_{n=1}^{\infty} \frac{1}{4^n} - \sum_{n=1}^{\infty} \frac{1}{(4 \cdot 3)^n} \right) \Rightarrow 1 + \frac{3}{2} \left( \sum_{n=1}^{\infty} \frac{1}{4^n} - \sum_{n=1}^{\infty} \frac{1}{12^n} \right) \]
Both of these components are geometric series:
\[ \sum_{n=1}^{\infty} \frac{1}{4^n} = \frac{\frac{1}{4}}{1 - \frac{1}{4}} = \frac{1}{3}, \quad \sum_{n=1}^{\infty} \frac{1}{12^n} = \frac{\frac{1}{12}}{1 - \frac{1}{12}} = \frac{1}{11} \]
Substitution into the expression yields:
\[ 1 + \frac{3}{2} \left( \frac{1}{3} - \frac{1}{11} \right) = 1 + \frac{3}{2} \cdot \frac{8}{33} = 1 + \frac{24}{66} = \frac{66 + 24}{66} = \frac{90}{66} = \frac{15}{11} \]
Correction: A review of the calculation indicates an error. The correct step is:
\[ 1 + \frac{3}{2} \left( \frac{1}{3} - \frac{1}{11} \right) = 1 + \frac{3}{2} \cdot \frac{11 - 3}{33} = 1 + \frac{3}{2} \cdot \frac{8}{33} = 1 + \frac{24}{66} \]
The subsequent simplification was:
\[ 1 + \frac{24}{66} = \frac{66}{66} + \frac{24}{66} = \frac{90}{66} = \frac{15}{11} \]
✅ Final Answer: \(\boxed{\frac{15}{11}}\)