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. 0
B. `4veca`
C. `5veca`
D. `6veca`

Answer

To solve the problem, we need to evaluate the expression \( \vec{AB} + 2\vec{BC} + 3\vec{CD} + 4\vec{DE} + 5\vec{EA} \) given that \( \vec{OA} = \vec{a} \).

### Step-by-Step Solution:

1. **Identify the Vectors**:
We know that \( O \) is the center of the pentagon, and we can express the vectors in terms of the position vectors of points \( A, B, C, D, \) and \( E \) relative to the center \( O \).
- Let \( \vec{A} = \vec{a} \)
- Let \( \vec{B} = \vec{b} \)
- Let \( \vec{C} = \vec{c} \)
- Let \( \vec{D} = \vec{d} \)
- Let \( \vec{E} = \vec{e} \)

2. **Express Each Segment Vector**:
- \( \vec{AB} = \vec{B} - \vec{A} = \vec{b} - \vec{a} \)
- \( \vec{BC} = \vec{C} - \vec{B} = \vec{c} - \vec{b} \)
- \( \vec{CD} = \vec{D} - \vec{C} = \vec{d} - \vec{c} \)
- \( \vec{DE} = \vec{E} - \vec{D} = \vec{e} - \vec{d} \)
- \( \vec{EA} = \vec{A} - \vec{E} = \vec{a} - \vec{e} \)

3. **Substitute into the Expression**:
Substitute these expressions 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. **Combine Like Terms**:
Now, let's combine all the terms:
\[ = \vec{b} - \vec{a} + 2\vec{c} - 2\vec{b} + 3\vec{d} - 3\vec{c} + 4\vec{e} - 4\vec{d} + 5\vec{a} - 5\vec{e} \]
Rearranging gives:
\[ = (-\vec{a} + 5\vec{a}) + (\vec{b} - 2\vec{b}) + (2\vec{c} - 3\vec{c}) + (3\vec{d} - 4\vec{d}) + (4\vec{e} - 5\vec{e}) \]
\[ = 4\vec{a} - \vec{b} - \vec{c} - \vec{d} - \vec{e} \]

5. **Sum of Vectors in a Regular Pentagon**:
In a regular pentagon, the sum of the position vectors from the center \( O \) to the vertices \( A, B, C, D, E \) is zero:
\[ \vec{a} + \vec{b} + \vec{c} + \vec{d} + \vec{e} = 0 \]
Therefore, we can substitute \( \vec{b} + \vec{c} + \vec{d} + \vec{e} = -\vec{a} \) into our expression:
\[ = 4\vec{a} - (-\vec{a}) = 4\vec{a} + \vec{a} = 5\vec{a} \]

6. **Final Result**:
Thus, we conclude that:
\[ \vec{AB} + 2\vec{BC} + 3\vec{CD} + 4\vec{DE} + 5\vec{EA} = 5\vec{a} \]

### Final Answer:
\[ \vec{AB} + 2\vec{BC} + 3\vec{CD} + 4\vec{DE} + 5\vec{EA} = 5\vec{a} \]