Sum of n terms of the series `1 + (1 + 3) + (1 + 3 + 5)+…` is

Options

A. `n^(2)`
B. `(n (n + 1) (2n + 1))/(6)`
C. `((n (n + 1))/(2))^(2)`
D. none

Answer

To find the sum of the first \( n \) terms of the series \( 1 + (1 + 3) + (1 + 3 + 5) + \ldots \), we can follow these steps:

### Step 1: Identify the Pattern
The series can be rewritten as:
- The 1st term: \( 1 \)
- The 2nd term: \( 1 + 3 = 4 \)
- The 3rd term: \( 1 + 3 + 5 = 9 \)
- The 4th term: \( 1 + 3 + 5 + 7 = 16 \)

We can observe that the \( n \)-th term of the series is the sum of the first \( n \) odd numbers.

### Step 2: Formula for the Sum of First \( n \) Odd Numbers
The sum of the first \( n \) odd numbers is given by the formula:
\[ T_n = n^2 \]
Thus, the \( n \)-th term of our series can be expressed as:
\[ T_n = n^2 \]

### Step 3: Find the Sum of the First \( n \) Terms
Now, we need to find the sum of the first \( n \) terms of the series:
\[ S_n = T_1 + T_2 + T_3 + \ldots + T_n \]
Substituting the values of \( T_n \):
\[ S_n = 1^2 + 2^2 + 3^2 + \ldots + n^2 \]

### Step 4: Use the Formula for the Sum of Squares
The formula for the sum of the squares of the first \( n \) natural numbers is:
\[ S_n = \frac{n(n + 1)(2n + 1)}{6} \]

### Final Answer
Thus, the sum of the first \( n \) terms of the series is:
\[ S_n = \frac{n(n + 1)(2n + 1)}{6} \]