STATEMENT-1 : If `1^(2) - 2^(2) + 3^(2) …….."to" 21 ` terms is 231
STATEMENT-2 : If ` 1^(3) - 2^(3) + 3^(3) - 4^(5) ………..` to 15 terms is 1856
STATEMENT-3 : If ` 1^(1) + 3^(2) + 5^(2) ……..` to 8 terms is 689

Options

A. F T F
B. T T F
C. T T T
D. F F T

Answer

To solve the given statements, we will analyze each statement one by one, verifying their correctness through calculations.

### Statement 1:
**Claim:** The sum \(1^2 - 2^2 + 3^2 - 4^2 + \ldots\) up to 21 terms equals 231.

**Solution:**
1. **Identify the pattern:** The series alternates between positive and negative squares. We can group the terms:
\[ S = (1^2) + (-2^2) + (3^2) + (-4^2) + \ldots + (21^2) \]
This can be rewritten as:
\[ S = (1^2 + 3^2 + 5^2 + \ldots + 21^2) - (2^2 + 4^2 + 6^2 + \ldots + 20^2) \]

2. **Calculate the sum of squares of odd numbers (up to 21):**
The sum of squares of the first \(n\) odd numbers is given by:
\[ \text{Sum of odd squares} = \frac{n(2n - 1)(2n + 1)}{3} \]
Here, \(n = 11\) (since there are 11 odd numbers from 1 to 21):
\[ \text{Sum of odd squares} = \frac{11(22)(23)}{3} = \frac{11 \times 506}{3} = 1852 \]

3. **Calculate the sum of squares of even numbers (up to 20):**
The sum of squares of the first \(n\) even numbers is given by:
\[ \text{Sum of even squares} = 2^2 \cdot \frac{n(n + 1)(2n + 1)}{6} \]
Here, \(n = 10\):
\[ \text{Sum of even squares} = 4 \cdot \frac{10(11)(21)}{6} = 4 \cdot 385 = 1540 \]

4. **Combine the results:**
\[ S = 1852 - 1540 = 312 \]
Since \(312 \neq 231\), **Statement 1 is false.**

### Statement 2:
**Claim:** The sum \(1^3 - 2^3 + 3^3 - 4^3 + \ldots\) up to 15 terms equals 1856.

**Solution:**
1. **Identify the pattern:** The series can be grouped similarly:
\[ S = (1^3 + 3^3 + 5^3 + \ldots + 15^3) - (2^3 + 4^3 + 6^3 + \ldots + 14^3) \]

2. **Calculate the sum of cubes of odd numbers (up to 15):**
The sum of cubes of the first \(n\) odd numbers is given by:
\[ \left(\frac{n(2n - 1)(2n + 1)}{3}\right)^2 \]
Here, \(n = 8\):
\[ \text{Sum of odd cubes} = \left(\frac{8(15)(17)}{3}\right)^2 = \left(680\right)^2 = 462400 \]

3. **Calculate the sum of cubes of even numbers (up to 14):**
The sum of cubes of the first \(n\) even numbers is given by:
\[ \left(\frac{n(n + 1)}{2}\right)^2 \]
Here, \(n = 7\):
\[ \text{Sum of even cubes} = \left(\frac{7(8)}{2}\right)^2 = 784 \]

4. **Combine the results:**
\[ S = 462400 - 784 = 461616 \]
Since \(461616 \neq 1856\), **Statement 2 is false.**

### Statement 3:
**Claim:** The sum \(1^2 + 3^2 + 5^2 + \ldots\) up to 8 terms equals 689.

**Solution:**
1. **Identify the pattern:** The series consists of the squares of the first 8 odd numbers:
\[ S = 1^2 + 3^2 + 5^2 + 7^2 + 9^2 + 11^2 + 13^2 + 15^2 \]

2. **Calculate the sum of squares of odd numbers:**
Using the formula for the sum of squares of the first \(n\) odd numbers:
\[ S = \frac{n(2n - 1)(2n + 1)}{3} \]
Here, \(n = 8\):
\[ S = \frac{8(15)(17)}{3} = \frac{2040}{3} = 680 \]

3. **Since \(680 \neq 689\), Statement 3 is false.**

### Conclusion:
- **Statement 1:** False
- **Statement 2:** False
- **Statement 3:** False