Find sum to first n groups of : `(1+3+9+27) +….. `

Answer

To find the sum of the series \( S = (1 + 3 + 9 + 27 + \ldots) \) up to the first \( n \) terms, we can follow these steps:

### Step 1: Identify the series
The series can be expressed as:
\[ S_n = 1 + 3 + 9 + 27 + \ldots \]
We can observe that each term in the series can be represented as powers of 3:
- The first term is \( 3^0 = 1 \)
- The second term is \( 3^1 = 3 \)
- The third term is \( 3^2 = 9 \)
- The fourth term is \( 3^3 = 27 \)

Thus, the \( n \)-th term can be represented as:
\[ T_n = 3^{n-1} \]

### Step 2: Write the sum in summation notation
We can express the sum of the first \( n \) terms as:
\[ S_n = \sum_{k=1}^{n} 3^{k-1} \]

### Step 3: Use the formula for the sum of a geometric series
The series \( S_n \) is a geometric series where:
- The first term \( a = 1 \) (which is \( 3^0 \))
- The common ratio \( r = 3 \)

The formula for the sum of the first \( n \) terms of a geometric series is given by:
\[ S_n = a \frac{r^n - 1}{r - 1} \]

Substituting the values of \( a \) and \( r \):
\[ S_n = 1 \cdot \frac{3^n - 1}{3 - 1} \]
\[ S_n = \frac{3^n - 1}{2} \]

### Step 4: Final expression for the sum
Thus, the sum of the first \( n \) terms of the series is:
\[ S_n = \frac{3^n - 1}{2} \]

### Summary
The required sum of the first \( n \) terms of the series \( (1 + 3 + 9 + 27 + \ldots) \) is:
\[ \frac{3^n - 1}{2} \]

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