Write the first five terms of each of the following sequences whose nth terms are:
(i) `a_(n)=2^(n)`
(ii) `T_(n)=(n(n+1)(2n+1))/6`

Answer

To find the first five terms of the given sequences, we will evaluate the nth terms provided.

### (i) For the sequence defined by \( a_n = 2^n \):

1. **Calculate \( a_1 \)**:
\[ a_1 = 2^1 = 2 \]

2. **Calculate \( a_2 \)**:
\[ a_2 = 2^2 = 4 \]

3. **Calculate \( a_3 \)**:
\[ a_3 = 2^3 = 8 \]

4. **Calculate \( a_4 \)**:
\[ a_4 = 2^4 = 16 \]

5. **Calculate \( a_5 \)**:
\[ a_5 = 2^5 = 32 \]

Thus, the first five terms of the sequence \( a_n = 2^n \) are:
\[ \text{First five terms: } 2, 4, 8, 16, 32 \]

### (ii) For the sequence defined by \( T_n = \frac{n(n+1)(2n+1)}{6} \):

1. **Calculate \( T_1 \)**:
\[ T_1 = \frac{1(1+1)(2 \cdot 1 + 1)}{6} = \frac{1 \cdot 2 \cdot 3}{6} = \frac{6}{6} = 1 \]

2. **Calculate \( T_2 \)**:
\[ T_2 = \frac{2(2+1)(2 \cdot 2 + 1)}{6} = \frac{2 \cdot 3 \cdot 5}{6} = \frac{30}{6} = 5 \]

3. **Calculate \( T_3 \)**:
\[ T_3 = \frac{3(3+1)(2 \cdot 3 + 1)}{6} = \frac{3 \cdot 4 \cdot 7}{6} = \frac{84}{6} = 14 \]

4. **Calculate \( T_4 \)**:
\[ T_4 = \frac{4(4+1)(2 \cdot 4 + 1)}{6} = \frac{4 \cdot 5 \cdot 9}{6} = \frac{180}{6} = 30 \]

5. **Calculate \( T_5 \)**:
\[ T_5 = \frac{5(5+1)(2 \cdot 5 + 1)}{6} = \frac{5 \cdot 6 \cdot 11}{6} = \frac{330}{6} = 55 \]

Thus, the first five terms of the sequence \( T_n = \frac{n(n+1)(2n+1)}{6} \) are:
\[ \text{First five terms: } 1, 5, 14, 30, 55 \]

### Summary of Results:
- For \( a_n = 2^n \): **2, 4, 8, 16, 32**
- For \( T_n = \frac{n(n+1)(2n+1)}{6} \): **1, 5, 14, 30, 55**