The sum of the following series:
`1.1^2 (1 - 0/1) + 2.2^(2) (1 - 1/2) + 3.3^2 (1 - 2/3) + 4.4^2 (1 - 3/4) +…..` upto n terms is :

Options

A. `n.n^2 (1 - 1/n)`
B. `(n(n +1)(2n + 1))/(6)`
C. `(n(n + 1)^2)/(4)`
D. none of these

Answer

To solve the series given in the question, we will break it down step by step.

### Step-by-Step Solution

1. **Understanding the Series**:
The series is given as:
\[ S = 1 \cdot 1^2 \left(1 - \frac{0}{1}\right) + 2 \cdot 2^2 \left(1 - \frac{1}{2}\right) + 3 \cdot 3^2 \left(1 - \frac{2}{3}\right) + 4 \cdot 4^2 \left(1 - \frac{3}{4}\right) + \ldots \]
We can express the general term for \( k \)-th term as:
\[ T_k = k \cdot k^2 \left(1 - \frac{k-1}{k}\right) \]

2. **Simplifying Each Term**:
The term simplifies as follows:
\[ T_k = k \cdot k^2 \left(1 - \frac{k-1}{k}\right) = k \cdot k^2 \left(\frac{1}{k}\right) = k^2 \]
Therefore, each term in the series is simply \( k^2 \).

3. **Writing the Series**:
The series can now be rewritten as:
\[ S = 1^2 + 2^2 + 3^2 + 4^2 + \ldots + n^2 \]

4. **Using the Formula for the Sum of Squares**:
The sum of the squares of the first \( n \) natural numbers is given by the formula:
\[ \text{Sum} = \frac{n(n + 1)(2n + 1)}{6} \]
Therefore, we can express \( S \) as:
\[ S = \frac{n(n + 1)(2n + 1)}{6} \]

5. **Final Result**:
Thus, the sum of the series up to \( n \) terms is:
\[ S = \frac{n(n + 1)(2n + 1)}{6} \]