If `P (n) = 2 + 4 + 6+…+ 2n, n in N` then ` P(k) = k (k +1) + 2 implies P (k +1 ) = (k +1 ) (k + 2)+ 2 AA k in N.` So, we can conciude that `P (n) = n (n +1) + 2` for

Options

A. all `n in N`
B. `n gt 1`
C. `n gt 2`
D. nothing can be said

Answer

To solve the problem, we need to evaluate the expression \( P(n) = 2 + 4 + 6 + \ldots + 2n \) and prove that it can be expressed as \( P(n) = n(n + 1) + 2 \).

### Step 1: Understand the series
The series \( P(n) = 2 + 4 + 6 + \ldots + 2n \) represents the sum of the first \( n \) even numbers.

### Step 2: Express the series
We can express the \( n \)-th even number as \( 2n \). Therefore, the series can be rewritten as:
\[ P(n) = 2(1 + 2 + 3 + \ldots + n) \]

### Step 3: Use the formula for the sum of the first \( n \) natural numbers
The sum of the first \( n \) natural numbers is given by the formula:
\[ 1 + 2 + 3 + \ldots + n = \frac{n(n + 1)}{2} \]
Substituting this into our expression for \( P(n) \):
\[ P(n) = 2 \cdot \frac{n(n + 1)}{2} \]

### Step 4: Simplify the expression
The \( 2 \) in the numerator and the \( 2 \) in the denominator cancel out:
\[ P(n) = n(n + 1) \]

### Step 5: Add the constant term
However, from the original question, we need to add \( 2 \) to this expression:
\[ P(n) = n(n + 1) + 2 \]

### Step 6: Verify the formula with specific values
To ensure our formula is correct, we can check it with small values of \( n \):

- For \( n = 1 \):
\[ P(1) = 2 \quad \text{and} \quad 1(1 + 1) + 2 = 1 \cdot 2 + 2 = 4 \quad \text{(Correct)} \]

- For \( n = 2 \):
\[ P(2) = 2 + 4 = 6 \quad \text{and} \quad 2(2 + 1) + 2 = 2 \cdot 3 + 2 = 6 + 2 = 8 \quad \text{(Incorrect)} \]

### Conclusion
Thus, we can conclude that:
\[ P(n) = n(n + 1) + 2 \]
is valid for \( n \) in natural numbers.