The value of `lim_(nrarroo)(1^(2)-2^(2)+3^(2)-4^(2)+5^(2)….+(2n+1)^(2))/(n^(2))` is equal to

Options

A. 2
B. 4
C. 6
D. 8

Answer

To solve the limit

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

we will break down the expression step by step.

### Step 1: Rewrite the numerator
The numerator consists of alternating squares. We can group the terms in pairs:

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

This can be rewritten as:

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

### Step 2: Simplify each pair
Using the difference of squares formula \(a^2 - b^2 = (a-b)(a+b)\), we can simplify each pair:

\[ 1^2 - 2^2 = (1-2)(1+2) = -1 \cdot 3 = -3, \]
\[ 3^2 - 4^2 = (3-4)(3+4) = -1 \cdot 7 = -7, \]
\[ 5^2 - 6^2 = (5-6)(5+6) = -1 \cdot 11 = -11, \]
and so on.

In general, for the \(k\)-th pair \((2k-1)^2 - (2k)^2\):

\[ (2k-1)^2 - (2k)^2 = -1 \cdot (4k - 1) = -(4k - 1). \]

### Step 3: Sum the pairs
The last term is \((2n+1)^2\). Therefore, the total sum of the numerator can be expressed as:

\[ -(3 + 7 + 11 + \ldots + (4n - 1)) + (2n + 1)^2. \]

### Step 4: Find the sum of the series
The series \(3, 7, 11, \ldots, (4n - 1)\) is an arithmetic series with the first term \(3\) and the last term \((4n - 1)\). The number of terms is \(n\).

The sum of an arithmetic series is given by:

\[ \text{Sum} = \frac{\text{Number of terms}}{2} \times (\text{First term} + \text{Last term}) = \frac{n}{2} \times (3 + (4n - 1)) = \frac{n}{2} \times (4n + 2) = n(2n + 1). \]

### Step 5: Substitute back into the limit
Now substituting back, we have:

\[ \text{Numerator} = -(n(2n + 1)) + (2n + 1)^2. \]

Calculating \((2n + 1)^2\):

\[ (2n + 1)^2 = 4n^2 + 4n + 1. \]

So the numerator becomes:

\[ -(n(2n + 1)) + (4n^2 + 4n + 1) = -2n^2 - n + 4n^2 + 4n + 1 = 2n^2 + 3n + 1. \]

### Step 6: Formulate the limit
Now we can write the limit as:

\[ \lim_{n \to \infty} \frac{2n^2 + 3n + 1}{n^2}. \]

### Step 7: Simplify the limit
Dividing each term by \(n^2\):

\[ \lim_{n \to \infty} \left(2 + \frac{3}{n} + \frac{1}{n^2}\right). \]

As \(n\) approaches infinity, \(\frac{3}{n}\) and \(\frac{1}{n^2}\) approach \(0\):

\[ \lim_{n \to \infty} \left(2 + 0 + 0\right) = 2. \]

### Final Answer
Thus, the value of the limit is:

\[ \boxed{2}. \]