Write the first six terms of each of following sequences,
(i) `a_(1)=-1,a_(n)=(a_(n-1))/n,(nge2)`
(ii) `a_(1)=4,a_(n+1)=2na_(n)`

Answer

To solve the problem, we need to find the first six terms of two sequences defined by their recurrence relations.

### (i) Sequence: \( a_1 = -1, a_n = \frac{a_{n-1}}{n} \) for \( n \geq 2 \)

1. **Calculate \( a_1 \)**:
\[ a_1 = -1 \]

2. **Calculate \( a_2 \)**:
\[ a_2 = \frac{a_1}{2} = \frac{-1}{2} = -\frac{1}{2} \]

3. **Calculate \( a_3 \)**:
\[ a_3 = \frac{a_2}{3} = \frac{-\frac{1}{2}}{3} = -\frac{1}{6} \]

4. **Calculate \( a_4 \)**:
\[ a_4 = \frac{a_3}{4} = \frac{-\frac{1}{6}}{4} = -\frac{1}{24} \]

5. **Calculate \( a_5 \)**:
\[ a_5 = \frac{a_4}{5} = \frac{-\frac{1}{24}}{5} = -\frac{1}{120} \]

6. **Calculate \( a_6 \)**:
\[ a_6 = \frac{a_5}{6} = \frac{-\frac{1}{120}}{6} = -\frac{1}{720} \]

**First six terms of the sequence**:
\[ a_1 = -1, \quad a_2 = -\frac{1}{2}, \quad a_3 = -\frac{1}{6}, \quad a_4 = -\frac{1}{24}, \quad a_5 = -\frac{1}{120}, \quad a_6 = -\frac{1}{720} \]

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### (ii) Sequence: \( a_1 = 4, a_{n+1} = 2n a_n \)

1. **Calculate \( a_1 \)**:
\[ a_1 = 4 \]

2. **Calculate \( a_2 \)**:
\[ a_2 = 2 \cdot 1 \cdot a_1 = 2 \cdot 1 \cdot 4 = 8 \]

3. **Calculate \( a_3 \)**:
\[ a_3 = 2 \cdot 2 \cdot a_2 = 2 \cdot 2 \cdot 8 = 32 \]

4. **Calculate \( a_4 \)**:
\[ a_4 = 2 \cdot 3 \cdot a_3 = 2 \cdot 3 \cdot 32 = 192 \]

5. **Calculate \( a_5 \)**:
\[ a_5 = 2 \cdot 4 \cdot a_4 = 2 \cdot 4 \cdot 192 = 1536 \]

6. **Calculate \( a_6 \)**:
\[ a_6 = 2 \cdot 5 \cdot a_5 = 2 \cdot 5 \cdot 1536 = 15360 \]

**First six terms of the sequence**:
\[ a_1 = 4, \quad a_2 = 8, \quad a_3 = 32, \quad a_4 = 192, \quad a_5 = 1536, \quad a_6 = 15360 \]

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### Summary of Results:
- For the first sequence:
\[ -1, -\frac{1}{2}, -\frac{1}{6}, -\frac{1}{24}, -\frac{1}{120}, -\frac{1}{720} \]

- For the second sequence:
\[ 4, 8, 32, 192, 1536, 15360 \]