For positive integers k=1,2,3,....n,, let `S_k` denotes the area of `triangleAOB_k` such that `angleAOB_k=(kpi)/(2n)`, OA=1 and `OB_k=k` The value of the `lim_(n->oo)1/n^2sum_(k-1)^nS_k` is

Options

A. `2/(pi^(2))`
B. `4/(pi^(2))`
C. `8/(pi^(2))`
D. `1/(2pi)^(2)`

Answer

To solve the problem, we need to find the limit:

\[ \lim_{n \to \infty} \frac{1}{n^2} \sum_{k=1}^{n} S_k \]

where \( S_k \) is the area of triangle \( AOB_k \) with \( OA = 1 \), \( OB_k = k \), and the angle \( \angle AOB_k = \frac{k \pi}{2n} \).

### Step 1: Calculate the Area \( S_k \)

The area \( S_k \) of triangle \( AOB_k \) can be calculated using the formula for the area of a triangle given two sides and the included angle:

\[ S_k = \frac{1}{2} \cdot OA \cdot OB_k \cdot \sin(\angle AOB_k) \]

Substituting the values of \( OA \) and \( OB_k \):

\[ S_k = \frac{1}{2} \cdot 1 \cdot k \cdot \sin\left(\frac{k \pi}{2n}\right) = \frac{k}{2} \sin\left(\frac{k \pi}{2n}\right) \]

### Step 2: Substitute \( S_k \) into the Summation

Now, we substitute \( S_k \) into the summation:

\[ \sum_{k=1}^{n} S_k = \sum_{k=1}^{n} \frac{k}{2} \sin\left(\frac{k \pi}{2n}\right) \]

This can be rewritten as:

\[ \sum_{k=1}^{n} S_k = \frac{1}{2} \sum_{k=1}^{n} k \sin\left(\frac{k \pi}{2n}\right) \]

### Step 3: Analyze the Limit

Now we need to analyze the limit:

\[ \lim_{n \to \infty} \frac{1}{n^2} \cdot \frac{1}{2} \sum_{k=1}^{n} k \sin\left(\frac{k \pi}{2n}\right) \]

We can express the sum as a Riemann sum. As \( n \to \infty \), the term \( \frac{k}{n} \) approaches \( x \) where \( x \) ranges from \( 0 \) to \( 1 \). Thus, we can rewrite the sum:

\[ \frac{1}{n^2} \sum_{k=1}^{n} k \sin\left(\frac{k \pi}{2n}\right) \approx \frac{1}{n^2} \cdot n^2 \cdot \int_0^1 x \sin\left(\frac{\pi x}{2}\right) \, dx \]

### Step 4: Evaluate the Integral

Now we need to compute the integral:

\[ \int_0^1 x \sin\left(\frac{\pi x}{2}\right) \, dx \]

Using integration by parts, let:

- \( u = x \) and \( dv = \sin\left(\frac{\pi x}{2}\right) dx \)
- Then \( du = dx \) and \( v = -\frac{2}{\pi} \cos\left(\frac{\pi x}{2}\right) \)

Now applying integration by parts:

\[ \int u \, dv = uv - \int v \, du \]

Calculating:

\[ \int_0^1 x \sin\left(\frac{\pi x}{2}\right) \, dx = \left[-\frac{2}{\pi} x \cos\left(\frac{\pi x}{2}\right)\right]_0^1 + \frac{2}{\pi} \int_0^1 \cos\left(\frac{\pi x}{2}\right) \, dx \]

Evaluating the boundary terms:

\[ = -\frac{2}{\pi} \cdot 1 \cdot \cos\left(\frac{\pi}{2}\right) + 0 = 0 \]

Now we compute the second integral:

\[ \int_0^1 \cos\left(\frac{\pi x}{2}\right) \, dx = \left[\frac{2}{\pi} \sin\left(\frac{\pi x}{2}\right)\right]_0^1 = \frac{2}{\pi} \cdot 1 - 0 = \frac{2}{\pi} \]

Putting it all together:

\[ \int_0^1 x \sin\left(\frac{\pi x}{2}\right) \, dx = 0 + \frac{2}{\pi} \cdot \frac{2}{\pi} = \frac{4}{\pi^2} \]

### Step 5: Final Limit Calculation

Now substituting back into our limit:

\[ \lim_{n \to \infty} \frac{1}{n^2} \cdot \frac{1}{2} \cdot n^2 \cdot \frac{4}{\pi^2} = \frac{2}{\pi^2} \]

Thus, the final answer is:

\[ \boxed{\frac{2}{\pi^2}} \]