Let P(n) be the statement :
`”n^(2)+n` is even"
Prove that P(n) is true for all `n in N` by Mathematical Induction

Answer

To prove that the statement \( P(n) : n^2 + n \) is even for all \( n \in \mathbb{N} \) using Mathematical Induction, we will follow the standard steps of induction: the base case, the induction hypothesis, and the induction step.

### Step 1: Base Case
We first check the base case, which is \( n = 1 \).

\[ P(1) : 1^2 + 1 = 1 + 1 = 2 \]

Since 2 is even, the base case holds true.

### Step 2: Induction Hypothesis
Next, we assume that the statement is true for some arbitrary natural number \( k \). That is, we assume:

\[ P(k) : k^2 + k \text{ is even} \]

This means there exists some integer \( \alpha \) such that:

\[ k^2 + k = 2\alpha \]

### Step 3: Induction Step
Now we need to prove that \( P(k + 1) \) is also true. We need to show that:

\[ P(k + 1) : (k + 1)^2 + (k + 1) \text{ is even} \]

Calculating \( P(k + 1) \):

\[ (k + 1)^2 + (k + 1) = (k^2 + 2k + 1) + (k + 1) \]

Simplifying this expression:

\[ = k^2 + 2k + 1 + k + 1 = k^2 + 3k + 2 \]

Now we can rewrite \( k^2 + 3k + 2 \) using our induction hypothesis. Recall that \( k^2 + k = 2\alpha \):

\[ k^2 + 3k + 2 = (k^2 + k) + 2k + 2 = 2\alpha + 2k + 2 \]

Factoring out a 2:

\[ = 2\alpha + 2(k + 1) = 2(\alpha + k + 1) \]

Since \( \alpha + k + 1 \) is an integer, we can denote it as \( m \) (where \( m = \alpha + k + 1 \)). Thus, we have:

\[ (k + 1)^2 + (k + 1) = 2m \]

This shows that \( (k + 1)^2 + (k + 1) \) is even.

### Conclusion
Since we have shown that the base case is true and that if \( P(k) \) is true, then \( P(k + 1) \) is also true, by the principle of mathematical induction, we conclude that \( P(n) : n^2 + n \) is even for all \( n \in \mathbb{N} \).

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