A series is given in the form `(1) + (2+3+4)+ (5+6+7+8+9)+.....` Find the sum of the numbers in the rth bracket.

Options

A. `(n-1)^(3)+n^(3)`
B. `(n+1)^(3)+8n^(2)`
C. `((n+1)(n+2))/(6n)`
D. `(n+1)^(3)+n^(3)`

Answer

To find the sum of the numbers in the r-th bracket of the series given by \( (1) + (2+3+4) + (5+6+7+8+9) + \ldots \), we will follow these steps:

### Step 1: Identify the number of terms in the r-th bracket
The number of terms in the r-th bracket follows the pattern of odd numbers:
- 1st bracket has 1 term,
- 2nd bracket has 3 terms,
- 3rd bracket has 5 terms,
- and so on.

Thus, the number of terms in the r-th bracket is given by:
\[ \text{Number of terms} = 2r - 1 \]

### Step 2: Find the first term of the r-th bracket
To find the first term of the r-th bracket, we need to determine the sum of the number of terms in all previous brackets. The total number of terms in the first \( r-1 \) brackets can be calculated as:
\[ \text{Total terms in first } (r-1) \text{ brackets} = 1 + 3 + 5 + \ldots + (2(r-1) - 1) \]
This is the sum of the first \( r-1 \) odd numbers, which is known to be:
\[ (r-1)^2 \]
Thus, the first term of the r-th bracket is:
\[ \text{First term} = 1 + (r-1)^2 = r^2 - r + 1 \]

### Step 3: Calculate the sum of the numbers in the r-th bracket
The r-th bracket contains \( 2r - 1 \) terms, starting from \( r^2 - r + 1 \) and having a common difference of 1 (since the numbers are consecutive). The sum of an arithmetic series can be calculated using the formula:
\[ S_n = \frac{n}{2} \times (\text{first term} + \text{last term}) \]
where \( n \) is the number of terms.

The last term of the r-th bracket can be found as:
\[ \text{Last term} = \text{First term} + (2r - 2) = (r^2 - r + 1) + (2r - 2) = r^2 + r - 1 \]

Now, substituting into the sum formula:
\[ S_r = \frac{2r - 1}{2} \times \left((r^2 - r + 1) + (r^2 + r - 1)\right) \]
\[ = \frac{2r - 1}{2} \times (2r^2) \]
\[ = (2r - 1)r^2 \]

### Final Result
Thus, the sum of the numbers in the r-th bracket is:
\[ S_r = r^2(2r - 1) \]