Sum up 3+5`+`11 +29 + .... To n terms .

Answer

To find the sum of the series \(3 + 5 + 11 + 29 + \ldots\) up to \(n\) terms, we can follow these steps:

### Step 1: Identify the Pattern
First, we need to identify the pattern in the series. The given terms are:
- \(t_1 = 3\)
- \(t_2 = 5\)
- \(t_3 = 11\)
- \(t_4 = 29\)

Let's analyze the differences between consecutive terms:
- \(5 - 3 = 2\)
- \(11 - 5 = 6\)
- \(29 - 11 = 18\)

The differences are \(2, 6, 18\). Now, let's analyze the ratios of these differences:
- \(6 / 2 = 3\)
- \(18 / 6 = 3\)

This suggests that the differences are multiplying by 3, indicating a geometric progression in the differences.

### Step 2: Generalize the \(n\)th Term
From the pattern observed, we can express the \(n\)th term \(t_n\) as:
\[ t_n = 3 + 2 \cdot (3^{n-1} - 1) \]
This simplifies to:
\[ t_n = 3^{n} - 1 \]

### Step 3: Sum of the Series
The sum \(S_n\) of the first \(n\) terms can be expressed as:
\[ S_n = \sum_{r=1}^{n} t_r = \sum_{r=1}^{n} (3^r - 1) \]
This can be split into two separate sums:
\[ S_n = \sum_{r=1}^{n} 3^r - \sum_{r=1}^{n} 1 \]

### Step 4: Calculate Each Sum
1. The sum of the first \(n\) terms of a geometric series \(3^r\) is given by:
\[ \sum_{r=1}^{n} 3^r = 3 \frac{3^n - 1}{3 - 1} = \frac{3^{n+1} - 3}{2} \]

2. The sum of \(n\) ones is simply:
\[ \sum_{r=1}^{n} 1 = n \]

### Step 5: Combine the Results
Putting it all together, we have:
\[ S_n = \frac{3^{n+1} - 3}{2} - n \]
This simplifies to:
\[ S_n = \frac{3^{n+1} - 3 - 2n}{2} \]

### Final Answer
Thus, the sum of the series \(3 + 5 + 11 + 29 + \ldots\) up to \(n\) terms is:
\[ S_n = \frac{3^{n+1} - 3 - 2n}{2} \]