Let O be the centre of a regular pentagon ABCDE and `vec(OA) = veca`, then `vec(AB) +vec(2BC) + vec(3CD) + vec(4DE) + vec(5EA)` is equals:

Options

A. `6veca`
B. `5veca`
C. `4veca`
D. `vec0`

Answer

To solve the problem, we need to evaluate the expression \( \vec{AB} + 2\vec{BC} + 3\vec{CD} + 4\vec{DE} + 5\vec{EA} \) in terms of the vector \( \vec{OA} \) where \( O \) is the center of the regular pentagon \( ABCDE \).

### Step-by-Step Solution:

1. **Understanding the Vectors**:
- Let \( \vec{OA} = \vec{a} \).
- The vectors from the center \( O \) to the vertices \( A, B, C, D, E \) can be represented as \( \vec{A}, \vec{B}, \vec{C}, \vec{D}, \vec{E} \) respectively.

2. **Expressing Each Segment**:
- The vector \( \vec{AB} \) can be expressed as:
\[ \vec{AB} = \vec{B} - \vec{A} \]
- The vector \( \vec{BC} \) can be expressed as:
\[ \vec{BC} = \vec{C} - \vec{B} \]
- The vector \( \vec{CD} \) can be expressed as:
\[ \vec{CD} = \vec{D} - \vec{C} \]
- The vector \( \vec{DE} \) can be expressed as:
\[ \vec{DE} = \vec{E} - \vec{D} \]
- The vector \( \vec{EA} \) can be expressed as:
\[ \vec{EA} = \vec{A} - \vec{E} \]

3. **Substituting into the Expression**:
- Substitute the expressions for each segment into the original equation:
\[ \vec{AB} + 2\vec{BC} + 3\vec{CD} + 4\vec{DE} + 5\vec{EA} = (\vec{B} - \vec{A}) + 2(\vec{C} - \vec{B}) + 3(\vec{D} - \vec{C}) + 4(\vec{E} - \vec{D}) + 5(\vec{A} - \vec{E}) \]

4. **Combining Like Terms**:
- Rearranging the terms gives:
\[ = \vec{B} - \vec{A} + 2\vec{C} - 2\vec{B} + 3\vec{D} - 3\vec{C} + 4\vec{E} - 4\vec{D} + 5\vec{A} - 5\vec{E} \]
- Combine the coefficients of each vector:
\[ = (-\vec{A} + 5\vec{A}) + (\vec{B} - 2\vec{B}) + (2\vec{C} - 3\vec{C}) + (3\vec{D} - 4\vec{D}) + (4\vec{E} - 5\vec{E}) \]
- This simplifies to:
\[ = 4\vec{A} - \vec{B} - \vec{C} - \vec{D} - \vec{E} \]

5. **Using the Property of Regular Pentagon**:
- In a regular pentagon, the sum of the position vectors of the vertices from the center is zero:
\[ \vec{A} + \vec{B} + \vec{C} + \vec{D} + \vec{E} = 0 \]
- Therefore, \( \vec{B} + \vec{C} + \vec{D} + \vec{E} = -\vec{A} \).

6. **Final Simplification**:
- Substitute this back into our expression:
\[ = 4\vec{A} - (-\vec{A}) = 4\vec{A} + \vec{A} = 5\vec{A} \]

### Conclusion:
Thus, we conclude that:
\[ \vec{AB} + 2\vec{BC} + 3\vec{CD} + 4\vec{DE} + 5\vec{EA} = 5\vec{A} \]