The sum of the series `1 + 2 xx 3 + 3 xx 5 + 4 xx 7+`...upto 11th term is

Options

A. 945
B. 946
C. 916
D. 915

Explanation

To find the sum of the series \( S = 1 + 2 \times 3 + 3 \times 5 + 4 \times 7 + \ldots \) up to the 11th term, we first need to identify the general term of the series.

### Step 1: Identify the nth term
The nth term of the series can be expressed as:
\[ T_n = n \times (2n - 1) \]
This is because:
- For \( n = 1 \): \( T_1 = 1 \times (2 \times 1 - 1) = 1 \)
- For \( n = 2 \): \( T_2 = 2 \times (2 \times 2 - 1) = 2 \times 3 = 6 \)
- For \( n = 3 \): \( T_3 = 3 \times (2 \times 3 - 1) = 3 \times 5 = 15 \)
- For \( n = 4 \): \( T_4 = 4 \times (2 \times 4 - 1) = 4 \times 7 = 28 \)
- And so on...

### Step 2: Write the sum of the series
The sum of the first 11 terms can be expressed as:
\[ S = \sum_{n=1}^{11} T_n = \sum_{n=1}^{11} n(2n - 1) \]

### Step 3: Expand the summation
We can expand the summation:
\[ S = \sum_{n=1}^{11} (2n^2 - n) = 2 \sum_{n=1}^{11} n^2 - \sum_{n=1}^{11} n \]

### Step 4: Use formulas for summation
We use the formulas for the sums of the first n natural numbers and the sum of the squares of the first n natural numbers:
- Sum of the first n natural numbers:
\[ \sum_{k=1}^{n} k = \frac{n(n + 1)}{2} \]
- Sum of the squares of the first n natural numbers:
\[ \sum_{k=1}^{n} k^2 = \frac{n(n + 1)(2n + 1)}{6} \]

### Step 5: Calculate the sums for n = 11
For \( n = 11 \):
- Calculate \( \sum_{n=1}^{11} n^2 \):
\[ \sum_{n=1}^{11} n^2 = \frac{11(11 + 1)(2 \times 11 + 1)}{6} = \frac{11 \times 12 \times 23}{6} = 506 \]
- Calculate \( \sum_{n=1}^{11} n \):
\[ \sum_{n=1}^{11} n = \frac{11(11 + 1)}{2} = \frac{11 \times 12}{2} = 66 \]

### Step 6: Substitute back into the sum
Now substitute these values back into the expression for \( S \):
\[ S = 2 \times 506 - 66 = 1012 - 66 = 946 \]

### Final Answer
Thus, the sum of the series up to the 11th term is:
\[ \boxed{946} \]