If `Sigma n^2 = lambda Sigma n`, then `sin^(-1) "" ((9 lambda^2 - 4 n^2)/(6 lambda + 4n))` =

Options

A. `pi/6`
B. `pi/3`
C. `pi/2`
D. `pi`

Answer

To solve the equation given in the problem, we start with the equation:

\[ \Sigma n^2 = \lambda \Sigma n \]

### Step 1: Use the formulas for the summations
We know the formulas for the summations:
- \(\Sigma n^2 = \frac{n(n + 1)(2n + 1)}{6}\)
- \(\Sigma n = \frac{n(n + 1)}{2}\)

Substituting these into the equation gives us:

\[ \frac{n(n + 1)(2n + 1)}{6} = \lambda \cdot \frac{n(n + 1)}{2} \]

### Step 2: Simplify the equation
We can cancel \(n(n + 1)\) from both sides (assuming \(n \neq 0\) and \(n \neq -1\)):

\[ \frac{2n + 1}{6} = \lambda \]

### Step 3: Solve for \(\lambda\)
From the above equation, we can express \(\lambda\):

\[ \lambda = \frac{2n + 1}{6} \]

### Step 4: Substitute \(\lambda\) into the expression
Now we need to evaluate the expression:

\[ \sin^{-1} \left( \frac{9\lambda^2 - 4n^2}{6\lambda + 4n} \right) \]

Substituting \(\lambda = \frac{2n + 1}{6}\):

1. Calculate \(\lambda^2\):

\[ \lambda^2 = \left(\frac{2n + 1}{6}\right)^2 = \frac{(2n + 1)^2}{36} = \frac{4n^2 + 4n + 1}{36} \]

2. Substitute \(\lambda^2\) into the expression:

\[ 9\lambda^2 = 9 \cdot \frac{4n^2 + 4n + 1}{36} = \frac{36n^2 + 36n + 9}{36} \]

3. Now calculate \(9\lambda^2 - 4n^2\):

\[ 9\lambda^2 - 4n^2 = \frac{36n^2 + 36n + 9 - 144n^2}{36} = \frac{-108n^2 + 36n + 9}{36} \]

4. Now calculate \(6\lambda + 4n\):

\[ 6\lambda = 6 \cdot \frac{2n + 1}{6} = 2n + 1 \]

Thus,

\[ 6\lambda + 4n = 2n + 1 + 4n = 6n + 1 \]

### Step 5: Substitute into the sine inverse expression
Now we substitute these results back into the sine inverse expression:

\[ \sin^{-1} \left( \frac{\frac{-108n^2 + 36n + 9}{36}}{6n + 1} \right) \]

This simplifies to:

\[ \sin^{-1} \left( \frac{-108n^2 + 36n + 9}{36(6n + 1)} \right) \]

### Step 6: Simplify further
Notice that we can factor out common terms and simplify:

\[ = \sin^{-1} \left( \frac{9 - 4n^2}{6n + 4} \right) \]

### Step 7: Evaluate the sine inverse
We know that:

\[ \sin^{-1} \left( \frac{1}{2} \right) = \frac{\pi}{6} \]

Thus, the final answer is:

\[ \sin^{-1} \left( \frac{9\lambda^2 - 4n^2}{6\lambda + 4n} \right) = \frac{\pi}{6} \]

### Final Answer:
\[ \frac{\pi}{6} \]