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find the limits of the numerical sequence

The given expression involves a sequence with a limit as \( n \to \infty \):

\[
\lim_{n \to \infty} \left( \frac{3}{4} + \frac{5}{16} + \frac{9}{64} + \dots + \left( \frac{1}{4^n} + \frac{2^n}{4^n} \right) \right).
\]

Let's break this down step by step:

1. The terms appear to follow a pattern:
   - First term: \( \frac{3}{4} \)
   - Second term: \( \frac{5}{16} \)
   - Third term: \( \frac{9}{64} \)
   - General term: \( \frac{k^2}{4^k} \), where \( k \) is the position of the term (for \( k \geq 1 \)).

2. The limit expression includes an additional term at the end, which is the sum of:
   \[
   \frac{1}{4^n} + \frac{2^n}{4^n} = \frac{1}{4^n} + \left( \frac{2}{4} \right)^n = \frac{1}{4^n} + \frac{2^n}{4^n}.
   \]
   Simplifying the second term:
   \[
   \frac{2^n}{4^n} = \left( \frac{1}{2} \right)^n.
   \]

3. The entire sequence resembles a series where each term has the form \( \frac{k^2}{4^k} \) plus the remaining terms involving powers of \( \frac{1}{4} \) and \( \frac{1}{2} \).

#### Limit Calculation:

The series involves a combination of decreasing terms as \( n \to \infty \). Let's analyze the behavior:

- The terms \( \frac{k^2}{4^k} \) decrease rapidly as \( k \to \infty \) because \( 4^k \) grows much faster than \( k^2 \), making each successive term smaller.
- The additional terms \( \frac{1}{4^n} \) and \( \frac{2^n}{4^n} \) tend to 0 as \( n \to \infty \), since both powers of \( 4 \) and \( \frac{1}{2} \) decay to 0.

Therefore, the entire expression converges to a limit, and since all terms decay to 0, the limit of the sequence is:

\[
\boxed{1}.
\]

Would you like more details or have any questions?

### Related Questions:
1. How do we prove the convergence of the series \( \sum_{k=1}^{\infty} \frac{k^2}{4^k} \)?
2. What is the behavior of the general term \( \frac{k^2}{4^k} \) as \( k \to \infty \)?
3. Can the series be summed using any special formulas for power series?
4. How does the inclusion of \( \frac{1}{4^n} + \frac{2^n}{4^n} \) affect the convergence?
5. What techniques can be used to find limits of similar sequences involving exponential decay?

### Tip:
For sequences involving powers, always check the rate at which terms decay, especially for powers like \( 4^n \) or \( 2^n \), as they often lead to zero in limits.

Find the limits of the numerical sequence as n approaches infinity.

Explore how to find the limit of a numerical sequence as n approaches infinity. This solution breaks down a complex series involving exponential decay and powers of 4, providing a step-by-step explanation suitable for advanced high school and early college students.

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