Write the first five terms of each of the sequences in Questions 1 to 6 whose nth terms are :
`a_(n)=n.(n^(2)+5)/(4)`

Explanation

To find the first five terms of the sequence defined by the nth term \( a_n = \frac{n(n^2 + 5)}{4} \), we will substitute \( n = 1, 2, 3, 4, 5 \) into the formula and calculate each term step by step.

### Step 1: Calculate \( a_1 \)
Substituting \( n = 1 \):
\[ a_1 = \frac{1(1^2 + 5)}{4} \]
Calculating inside the parentheses:
\[ 1^2 + 5 = 1 + 5 = 6 \]
Now substituting back:
\[ a_1 = \frac{1 \cdot 6}{4} = \frac{6}{4} = \frac{3}{2} \]

### Step 2: Calculate \( a_2 \)
Substituting \( n = 2 \):
\[ a_2 = \frac{2(2^2 + 5)}{4} \]
Calculating inside the parentheses:
\[ 2^2 + 5 = 4 + 5 = 9 \]
Now substituting back:
\[ a_2 = \frac{2 \cdot 9}{4} = \frac{18}{4} = \frac{9}{2} \]

### Step 3: Calculate \( a_3 \)
Substituting \( n = 3 \):
\[ a_3 = \frac{3(3^2 + 5)}{4} \]
Calculating inside the parentheses:
\[ 3^2 + 5 = 9 + 5 = 14 \]
Now substituting back:
\[ a_3 = \frac{3 \cdot 14}{4} = \frac{42}{4} = \frac{21}{2} \]

### Step 4: Calculate \( a_4 \)
Substituting \( n = 4 \):
\[ a_4 = \frac{4(4^2 + 5)}{4} \]
Calculating inside the parentheses:
\[ 4^2 + 5 = 16 + 5 = 21 \]
Now substituting back:
\[ a_4 = \frac{4 \cdot 21}{4} = 21 \]

### Step 5: Calculate \( a_5 \)
Substituting \( n = 5 \):
\[ a_5 = \frac{5(5^2 + 5)}{4} \]
Calculating inside the parentheses:
\[ 5^2 + 5 = 25 + 5 = 30 \]
Now substituting back:
\[ a_5 = \frac{5 \cdot 30}{4} = \frac{150}{4} = \frac{75}{2} \]

### Summary of the First Five Terms
The first five terms of the sequence are:
\[ a_1 = \frac{3}{2}, \quad a_2 = \frac{9}{2}, \quad a_3 = \frac{21}{2}, \quad a_4 = 21, \quad a_5 = \frac{75}{2} \]

### Final Answer
The first five terms are:
\[ \frac{3}{2}, \frac{9}{2}, \frac{21}{2}, 21, \frac{75}{2} \]