Sum of the series `1+2.3+3.5+4.7+........` up to `11` terms is

Options

A. 942
B. 892
C. 946
D. 960

Answer

To find the sum of the series \( S = 1 + 2 \cdot 3 + 3 \cdot 5 + 4 \cdot 7 + \ldots \) up to 11 terms, we can express the general term of the series and then sum it up.

### Step-by-Step Solution:

1. **Identify the General Term**:
The series can be rewritten as:
\[ S = 1 \cdot 1 + 2 \cdot 3 + 3 \cdot 5 + 4 \cdot 7 + \ldots \]
The first term is \( r \) and the second term can be observed as \( 2r - 1 \) (which gives the odd numbers). Therefore, the general term \( T_r \) can be expressed as:
\[ T_r = r \cdot (2r - 1) \]

2. **Sum the General Terms**:
We need to sum this from \( r = 1 \) to \( r = 11 \):
\[ S = \sum_{r=1}^{11} T_r = \sum_{r=1}^{11} r(2r - 1) \]
Expanding this gives:
\[ S = \sum_{r=1}^{11} (2r^2 - r) \]
This can be separated into two sums:
\[ S = 2 \sum_{r=1}^{11} r^2 - \sum_{r=1}^{11} r \]

3. **Use the Formulas for Summation**:
We use the following formulas:
- The sum of the first \( n \) natural numbers:
\[ \sum_{r=1}^{n} r = \frac{n(n + 1)}{2} \]
- The sum of the squares of the first \( n \) natural numbers:
\[ \sum_{r=1}^{n} r^2 = \frac{n(n + 1)(2n + 1)}{6} \]

For \( n = 11 \):
- Calculate \( \sum_{r=1}^{11} r^2 \):
\[ \sum_{r=1}^{11} r^2 = \frac{11 \cdot 12 \cdot 23}{6} = \frac{3036}{6} = 506 \]
- Calculate \( \sum_{r=1}^{11} r \):
\[ \sum_{r=1}^{11} r = \frac{11 \cdot 12}{2} = 66 \]

4. **Substitute Back into the Sum**:
Now substitute these values back into the expression for \( S \):
\[ S = 2 \cdot 506 - 66 \]
\[ S = 1012 - 66 = 946 \]

### Final Answer:
The sum of the series up to 11 terms is \( \boxed{946} \).