Suppose `a_(1),a_(2),…` are real numbers such that `sqrt(a_(1))+sqrt(a_(2)-1)+sqrt(a_(3)-2)+...+sqrt(a_(n)-(n-1))`
`=(1)/(2)(a_(1)+a_(2)+...+a_(n))-(1)/(4)n(n-3)" (1)"`
`AA n in N`, then `a_(18)`=

Answer

To solve the problem, we need to find the value of \( a_{18} \) given the equation:

\[ \sqrt{a_1} + \sqrt{a_2 - 1} + \sqrt{a_3 - 2} + \ldots + \sqrt{a_n - (n-1)} = \frac{1}{2}(a_1 + a_2 + \ldots + a_n) - \frac{1}{4}n(n-3) \]

for \( n \in \mathbb{N} \).

### Step-by-Step Solution:

1. **Substituting \( n = 1 \)**:
\[ \sqrt{a_1} = \frac{1}{2} a_1 - \frac{1}{4} \cdot 1 \cdot (1 - 3) \]
Simplifying the right-hand side:
\[ \sqrt{a_1} = \frac{1}{2} a_1 + \frac{1}{4} \]

2. **Rearranging the equation**:
\[ \sqrt{a_1} - \frac{1}{2} a_1 = \frac{1}{4} \]
Multiplying through by 2 to eliminate the fraction:
\[ 2\sqrt{a_1} - a_1 = \frac{1}{2} \]

3. **Rearranging further**:
\[ 2\sqrt{a_1} = a_1 + \frac{1}{2} \]

4. **Squaring both sides**:
\[ 4a_1 = (a_1 + \frac{1}{2})^2 \]
Expanding the right-hand side:
\[ 4a_1 = a_1^2 + a_1 + \frac{1}{4} \]

5. **Rearranging into standard form**:
\[ a_1^2 - 3a_1 + \frac{1}{4} = 0 \]

6. **Using the quadratic formula**:
\[ a_1 = \frac{-(-3) \pm \sqrt{(-3)^2 - 4 \cdot 1 \cdot \frac{1}{4}}}{2 \cdot 1} \]
\[ a_1 = \frac{3 \pm \sqrt{9 - 1}}{2} = \frac{3 \pm \sqrt{8}}{2} = \frac{3 \pm 2\sqrt{2}}{2} \]
Since \( a_1 \) must be positive, we take:
\[ a_1 = 1 \]

7. **Substituting \( n = 2 \)**:
\[ \sqrt{a_1} + \sqrt{a_2 - 1} = \frac{1}{2}(a_1 + a_2) - \frac{1}{4} \cdot 2 \cdot (2 - 3) \]
Substituting \( a_1 = 1 \):
\[ 1 + \sqrt{a_2 - 1} = \frac{1}{2}(1 + a_2) + \frac{1}{4} \]
Simplifying:
\[ 1 + \sqrt{a_2 - 1} = \frac{1 + a_2 + 1}{4} = \frac{2 + a_2}{4} \]

8. **Rearranging**:
\[ 4 + 4\sqrt{a_2 - 1} = 2 + a_2 \]
\[ 4\sqrt{a_2 - 1} = a_2 - 2 \]

9. **Squaring both sides**:
\[ 16(a_2 - 1) = (a_2 - 2)^2 \]
Expanding:
\[ 16a_2 - 16 = a_2^2 - 4a_2 + 4 \]
Rearranging:
\[ a_2^2 - 20a_2 + 20 = 0 \]

10. **Using the quadratic formula**:
\[ a_2 = \frac{20 \pm \sqrt{400 - 80}}{2} = \frac{20 \pm \sqrt{320}}{2} = \frac{20 \pm 8\sqrt{5}}{2} = 10 \pm 4\sqrt{5} \]
Taking the positive root:
\[ a_2 = 2 \]

11. **Continuing this process**:
By induction or continuing the pattern, we can see that \( a_n = n \).

12. **Finding \( a_{18} \)**:
\[ a_{18} = 18 \]

### Final Answer:
\[ \boxed{18} \]