Suppose `A_(1), A_(2), …, A_(5)` are vertices of a regular pentagon with O as centre.
If `sum_(i=1)^(4)(OA_(i) times OA_(i+1))=lambda(OA_(1) times OA_(2))`
then `lambda=`_______

Answer

To solve the problem, we need to analyze the expression given in the question. We have a regular pentagon with vertices \( A_1, A_2, A_3, A_4, A_5 \) and center \( O \). The expression we need to evaluate is:

\[ \sum_{i=1}^{4} (OA_i \times OA_{i+1}) = \lambda (OA_1 \times OA_2) \]

### Step 1: Understand the Geometry of the Pentagon
In a regular pentagon, all sides and angles are equal. The distance from the center \( O \) to any vertex \( A_i \) is the same, which we can denote as \( OA = k \) (where \( k \) is a constant).

### Step 2: Determine the Angles
The angle between any two consecutive radius vectors \( OA_i \) and \( OA_{i+1} \) is \( \frac{2\pi}{5} \) radians (or \( 72^\circ \)). This is because the total angle around point \( O \) is \( 2\pi \) radians, and there are 5 equal segments.

### Step 3: Calculate the Cross Products
The magnitude of the cross product \( OA_i \times OA_{i+1} \) can be calculated using the formula:
\[ |OA_i \times OA_{i+1}| = |OA_i| |OA_{i+1}| \sin(\theta) \]
Since \( |OA_i| = |OA_{i+1}| = k \) and \( \theta = \frac{2\pi}{5} \), we have:
\[ |OA_i \times OA_{i+1}| = k \cdot k \cdot \sin\left(\frac{2\pi}{5}\right) = k^2 \sin\left(\frac{2\pi}{5}\right) \]

### Step 4: Evaluate the Summation
Now, we need to evaluate the left-hand side:
\[ \sum_{i=1}^{4} (OA_i \times OA_{i+1}) = OA_1 \times OA_2 + OA_2 \times OA_3 + OA_3 \times OA_4 + OA_4 \times OA_5 \]
Since each term has the same magnitude, we can express this as:
\[ = 4 (OA_1 \times OA_2) \]
Thus, the left-hand side becomes:
\[ 4 k^2 \sin\left(\frac{2\pi}{5}\right) \]

### Step 5: Right-Hand Side
The right-hand side is given as:
\[ \lambda (OA_1 \times OA_2) = \lambda (k^2 \sin\left(\frac{2\pi}{5}\right)) \]

### Step 6: Equate Both Sides
Now, we equate the left-hand side and the right-hand side:
\[ 4 k^2 \sin\left(\frac{2\pi}{5}\right) = \lambda (k^2 \sin\left(\frac{2\pi}{5}\right)) \]

### Step 7: Solve for \( \lambda \)
Assuming \( k^2 \sin\left(\frac{2\pi}{5}\right) \neq 0 \), we can divide both sides by \( k^2 \sin\left(\frac{2\pi}{5}\right) \):
\[ \lambda = 4 \]

### Final Answer
Thus, the value of \( \lambda \) is:
\[ \lambda = 4 \]