The sum of the `n` terms of the series `1+(1+3)+(1+3+5)+...` is

Options

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

Explanation

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 general term of the series
The series can be expressed as:
- The first term is \(1\).
- The second term is \(1 + 3\).
- The third term is \(1 + 3 + 5\).

We can see that each term is the sum of the first \(k\) odd numbers, where \(k\) is the term number. The \(k\)-th odd number can be expressed as \(2k - 1\).

### Step 2: Find the sum of the first \(k\) odd numbers
The sum of the first \(k\) odd numbers is given by the formula:
\[ T_k = 1 + 3 + 5 + \ldots + (2k - 1) = k^2 \]
Thus, the \(k\)-th term of the series can be represented as:
\[ T_k = k^2 \]

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

### 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:
\[ \sum_{k=1}^{n} k^2 = \frac{n(n + 1)(2n + 1)}{6} \]

### Step 5: Final expression for the sum of the series
Thus, the sum of the first \(n\) terms of the series is:
\[ S_n = \frac{n(n + 1)(2n + 1)}{6} \]

### Conclusion
The sum of the first \(n\) terms of the series \(1 + (1 + 3) + (1 + 3 + 5) + \ldots\) is:
\[ \frac{n(n + 1)(2n + 1)}{6} \]