Let `P(n) = 1 + 3 + 5 + ... + (2n - 1) = 3 + n^2`, then which of the following is true?
(i) P (3) is correct
(ii) P (2) is correct
(iii) `P(m) rArr p(m+1)`
(iv) `P(m) cancel(rArr) P(m+1)`

Options

A. P (3) is correct
B. P (2) is correct
C. `P(m) rArr p(m+1)`
D. `P(m) cancel(rArr) P(m+1)`

Answer

To solve the problem, we need to verify the statements regarding the function \( P(n) = 1 + 3 + 5 + \ldots + (2n - 1) = 3 + n^2 \).

### Step 1: Verify \( P(3) \)

**Left-hand side (LHS)**:
\[ P(3) = 1 + 3 + 5 = 9 \]

**Right-hand side (RHS)**:
\[ 3 + 3^2 = 3 + 9 = 12 \]

Since \( LHS \neq RHS \), we conclude that \( P(3) \) is false.

### Step 2: Verify \( P(2) \)

**Left-hand side (LHS)**:
\[ P(2) = 1 + 3 = 4 \]

**Right-hand side (RHS)**:
\[ 3 + 2^2 = 3 + 4 = 7 \]

Since \( LHS \neq RHS \), we conclude that \( P(2) \) is false.

### Step 3: Verify \( P(m) \implies P(m+1) \)

Assuming \( P(m) \) is true:
\[ P(m) = 1 + 3 + 5 + \ldots + (2m - 1) = 3 + m^2 \]

Now, we need to find \( P(m+1) \):
\[ P(m+1) = P(m) + (2m + 1) = (3 + m^2) + (2m + 1) = 3 + m^2 + 2m + 1 = 4 + m^2 + 2m \]

Now, we can rewrite \( 4 + m^2 + 2m \):
\[ 4 + m^2 + 2m = 3 + (m + 1)^2 \]

Thus, we have:
\[ P(m+1) = 3 + (m + 1)^2 \]

This shows that \( P(m) \implies P(m+1) \) is true.

### Step 4: Verify \( P(m) \cancel{\implies} P(m+1) \)

Since we have already established that \( P(m) \implies P(m+1) \) is true, the statement \( P(m) \cancel{\implies} P(m+1) \) is false.

### Conclusion

From our verification:
- (i) \( P(3) \) is false.
- (ii) \( P(2) \) is false.
- (iii) \( P(m) \implies P(m+1) \) is true.
- (iv) \( P(m) \cancel{\implies} P(m+1) \) is false.

Thus, the only true statement is (iii).