Find the sum of first n groups of (1) + (1+3) +(1+3+9) + (1+3+9 +27) + …

Options

A. a)`1/2 (3^n -1)`
B. b)`3/4 (3^n -1) - n/2`
C. c)`n^2 +1`
D. d)none of these

Answer

To find the sum of the first n groups of the series (1) + (1 + 3) + (1 + 3 + 9) + (1 + 3 + 9 + 27) + ..., we can break it down step by step.

### Step 1: Identify the pattern in the series
The series consists of groups where each group contains terms that are powers of 3:
- The first group is \( 1 = 3^0 \)
- The second group is \( 1 + 3 = 3^0 + 3^1 \)
- The third group is \( 1 + 3 + 9 = 3^0 + 3^1 + 3^2 \)
- The fourth group is \( 1 + 3 + 9 + 27 = 3^0 + 3^1 + 3^2 + 3^3 \)

Thus, the nth group can be expressed as:
\[ 1 + 3 + 3^2 + 3^3 + \ldots + 3^{n-1} \]

### Step 2: Recognize the series as a geometric progression (GP)
The nth group is a geometric series where:
- The first term \( a = 1 \)
- The common ratio \( r = 3 \)
- The number of terms \( n \)

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

### Step 3: Apply the formula to find the sum of the nth group
Substituting the values into the formula:
\[ S_n = \frac{1(3^n - 1)}{3 - 1} = \frac{3^n - 1}{2} \]

### Step 4: Find the total sum of the first n groups
Now, we need to find the sum of all the groups from 1 to n:
\[ \text{Total Sum} = S_1 + S_2 + S_3 + \ldots + S_n \]
Where \( S_k = \frac{3^k - 1}{2} \) for each group k.

### Step 5: Calculate the total sum
The total sum can be expressed as:
\[ \text{Total Sum} = \sum_{k=1}^{n} S_k = \sum_{k=1}^{n} \frac{3^k - 1}{2} \]
This can be separated into two sums:
\[ \text{Total Sum} = \frac{1}{2} \left( \sum_{k=1}^{n} 3^k - \sum_{k=1}^{n} 1 \right) \]
The second sum is simply \( n \) because we are summing 1, n times.

### Step 6: Calculate the sum of the geometric series
The sum \( \sum_{k=1}^{n} 3^k \) is also a geometric series:
\[ \sum_{k=1}^{n} 3^k = 3 \frac{3^n - 1}{3 - 1} = \frac{3(3^n - 1)}{2} \]

### Step 7: Combine the results
Now substituting back into the total sum:
\[ \text{Total Sum} = \frac{1}{2} \left( \frac{3(3^n - 1)}{2} - n \right) \]
Simplifying this:
\[ \text{Total Sum} = \frac{3(3^n - 1)}{4} - \frac{n}{2} \]

### Final Result
Thus, the sum of the first n groups is:
\[ \text{Total Sum} = \frac{3}{4}(3^n - 1) - \frac{n}{2} \]