If `lim_(n to oo) sum_(k=2)^(n)cos^(-1)((1+sqrt((k-1)(k+2)(k+1)k))/(k(k+1)))=(120pi)/(lambda)`, find the value of `lambda`.

Options

A. 700
B. 720
C. 760
D. None of these

Answer

To solve the problem step by step, we will analyze the limit and the summation involved in the expression.

### Step 1: Rewrite the Expression
We start with the limit:
\[ \lim_{n \to \infty} \sum_{k=2}^{n} \cos^{-1}\left(\frac{1 + \sqrt{(k-1)(k+2)(k+1)k}}{k(k+1)}\right) \]

### Step 2: Simplify the Argument of the Inverse Cosine
We can simplify the expression inside the inverse cosine:
\[ \sqrt{(k-1)(k+2)(k+1)k} = \sqrt{(k^2 + k - 2)(k^2 + 2k - 1)} \]
This can be rewritten as:
\[ \sqrt{(k^2 - 1)(k^2 + 2k)} = \sqrt{k^4 + k^2 - 2k^2 - 1} = \sqrt{k^4 + k^2 - 2} \]

### Step 3: Rewrite the Inverse Cosine
Now we rewrite the argument:
\[ \cos^{-1}\left(\frac{1 + \sqrt{(k-1)(k+2)(k+1)k}}{k(k+1)}\right) \]
This can be expressed as:
\[ \cos^{-1}\left(\frac{1}{k(k+1)} + \sqrt{\frac{(k-1)(k+2)(k+1)k}{k^2(k+1)^2}}\right) \]

### Step 4: Use the Cosine Addition Formula
Using the identity:
\[ \cos^{-1}(x) + \cos^{-1}(y) = \cos^{-1}(xy + \sqrt{(1-x^2)(1-y^2)}) \]
we can simplify further.

### Step 5: Recognize the Telescoping Series
The sum becomes:
\[ \sum_{k=2}^{n} \left(\cos^{-1}\left(\frac{1}{k+1}\right) - \cos^{-1}\left(\frac{1}{k}\right)\right) \]
This is a telescoping series where most terms cancel out.

### Step 6: Evaluate the Limit
After cancellation, we are left with:
\[ \cos^{-1}\left(\frac{1}{n+1}\right) - \cos^{-1}\left(\frac{1}{2}\right) \]
Taking the limit as \( n \to \infty \):
\[ \lim_{n \to \infty} \left(\cos^{-1}(0) - \cos^{-1}\left(\frac{1}{2}\right)\right) = \frac{\pi}{2} - \frac{\pi}{3} = \frac{\pi}{6} \]

### Step 7: Set the Equation
We have:
\[ \frac{\pi}{6} = \frac{120\pi}{\lambda} \]
Cross-multiplying gives:
\[ \lambda = 720 \]

### Final Answer
Thus, the value of \( \lambda \) is:
\[ \boxed{720} \]