Evaluate `lim_(ntooo) ((1^(2)-2^(2)+3^(2)-4^(2)+5^(2)+...n" terms"))/(n^(2)).`

Explanation

To evaluate the limit

\[ \lim_{n \to \infty} \frac{(1^2 - 2^2 + 3^2 - 4^2 + 5^2 + \ldots + (-1)^{n+1} n^2)}{n^2}, \]

we can start by rewriting the expression in the numerator. The series alternates between positive and negative squares of integers.

### Step 1: Group the terms in pairs

We can group the terms in pairs:

\[ (1^2 - 2^2) + (3^2 - 4^2) + (5^2 - 6^2) + \ldots \]

Each pair can be simplified as follows:

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

### Step 2: Determine the number of pairs

If \( n \) is even, we have \( n/2 \) pairs. If \( n \) is odd, we have \( (n-1)/2 \) pairs plus the last term \( n^2 \). However, for large \( n \), the last term \( n^2 \) will not significantly affect the limit.

### Step 3: Calculate the sum of pairs

For \( n \) even, the sum becomes:

\[ \sum_{k=1}^{n/2} (-2(2k-1) - 1) = -2\sum_{k=1}^{n/2} (2k-1) - \frac{n}{2}. \]

The sum of the first \( m \) odd numbers is \( m^2 \). Thus:

\[ \sum_{k=1}^{n/2} (2k-1) = \left(\frac{n}{2}\right)^2 = \frac{n^2}{4}. \]

So, we have:

\[ -2 \cdot \frac{n^2}{4} - \frac{n}{2} = -\frac{n^2}{2} - \frac{n}{2}. \]

### Step 4: Write the limit expression

Now, substituting back into the limit expression, we have:

\[ \lim_{n \to \infty} \frac{-\frac{n^2}{2} - \frac{n}{2}}{n^2}. \]

### Step 5: Simplify the limit

This can be simplified to:

\[ \lim_{n \to \infty} \left(-\frac{1}{2} - \frac{1}{2n}\right). \]

As \( n \to \infty \), the term \( \frac{1}{2n} \) approaches 0. Therefore, we get:

\[ -\frac{1}{2} - 0 = -\frac{1}{2}. \]

### Final Result

Thus, the limit evaluates to:

\[ \boxed{-\frac{1}{2}}. \]