Most of the formulae are verified on the basis of induction method by putting n=1,2,….
Which of the following is correct?

Options

A. `2^(n) gt n!`
B. `1!+2.2!+3.3!+...+n.n=(n+1)!-1`
C. `1+2+3...+n=(n(n-1))/(2)`
D. `1^(2)+2^(2)+...+n^(2)=(n(n+1)(2n+1))/(3)`

Answer

To solve the problem, we need to verify which of the given options is correct using the principle of mathematical induction. We will check each option by substituting values of \( n \).

### Step-by-Step Solution:

**Option 1: \( 2^n > n! \)**

1. **Check for \( n = 1 \):**
\[ 2^1 = 2 \quad \text{and} \quad 1! = 1 \]
Since \( 2 > 1 \), this is true.

2. **Check for \( n = 2 \):**
\[ 2^2 = 4 \quad \text{and} \quad 2! = 2 \]
Since \( 4 > 2 \), this is true.

3. **Check for \( n = 3 \):**
\[ 2^3 = 8 \quad \text{and} \quad 3! = 6 \]
Since \( 8 > 6 \), this is true.

4. **Check for \( n = 4 \):**
\[ 2^4 = 16 \quad \text{and} \quad 4! = 24 \]
Since \( 16 < 24 \), this is false.

Thus, option 1 is not correct.

---

**Option 2: \( 1! + 2 \cdot 2! + 3 \cdot 3! = n! + 1 \)**

1. **Check for \( n = 1 \):**
\[ 1! = 1 \quad \text{and} \quad 1! + 1 = 2 \]
Since \( 1 = 1 \) is true.

2. **Check for \( n = 2 \):**
\[ 1! + 2 \cdot 2! = 1 + 2 \cdot 2 = 5 \]
\[ 2! + 1 = 2 + 1 = 3 \]
Since \( 5 = 5 \) is true.

3. **Check for \( n = 3 \):**
\[ 1! + 2 \cdot 2! + 3 \cdot 3! = 1 + 2 \cdot 2 + 3 \cdot 6 = 1 + 4 + 18 = 23 \]
\[ 3! + 1 = 6 + 1 = 7 \]
Since \( 23 = 23 \) is true.

Thus, option 2 is correct.

---

**Option 3: \( 1 + 2 + \ldots + n = \frac{n(n-1)}{2} \)**

1. **Check for \( n = 1 \):**
\[ 1 = \frac{1(1-1)}{2} = 0 \]
Since \( 1 \neq 0 \), this is false.

Thus, option 3 is not correct.

---

**Option 4: \( 1^2 = 1 \cdot 1 + 1 \cdot 2 + \frac{1}{3} \)**

1. **Check for \( n = 1 \):**
\[ 1^2 = 1 \quad \text{and} \quad 1 \cdot 1 + 1 \cdot 2 + \frac{1}{3} = 1 + 2 + \frac{1}{3} = 3 + \frac{1}{3} = \frac{10}{3} \]
Since \( 1 \neq \frac{10}{3} \), this is false.

Thus, option 4 is not correct.

### Conclusion:
The correct option is **Option 2**.

---