Let `P(n) :n^(2)+n` is even.Is P(1) is true?

Answer

To determine whether \( P(1) \) is true, we first need to evaluate the expression given by \( P(n) \), which is defined as:

\[ P(n) : n^2 + n \]

We want to check if \( P(1) \) is even. Let's calculate \( P(1) \):

1. Substitute \( n = 1 \) into the expression:
\[ P(1) = 1^2 + 1 \]

2. Calculate \( 1^2 \):
\[ 1^2 = 1 \]

3. Now add \( 1 \) to the result:
\[ P(1) = 1 + 1 = 2 \]

4. Finally, we need to check if \( 2 \) is even:
\[ 2 \text{ is an even number.} \]

Since \( P(1) = 2 \) and \( 2 \) is even, we conclude that \( P(1) \) is true.

Thus, the answer is:
\[ \text{Yes, } P(1) \text{ is true.} \]