Write the first five terms of each of the following sequences whose nth terms are:
(i) `a_(n)=2n+5` (ii) `a_(n)=n(n+2)`
(iii) `a_(n)=(n-3)/4` (iv) `a_(n)=n/(n+1)`
(v) `a_(n)=(n(n^(2)+5))/4`

Answer

To find the first five terms of each sequence defined by the nth term formulas, we will substitute values of n from 1 to 5 into each formula.

### (i) \( a_n = 2n + 5 \)

1. For \( n = 1 \):
\[ a_1 = 2(1) + 5 = 2 + 5 = 7 \]

2. For \( n = 2 \):
\[ a_2 = 2(2) + 5 = 4 + 5 = 9 \]

3. For \( n = 3 \):
\[ a_3 = 2(3) + 5 = 6 + 5 = 11 \]

4. For \( n = 4 \):
\[ a_4 = 2(4) + 5 = 8 + 5 = 13 \]

5. For \( n = 5 \):
\[ a_5 = 2(5) + 5 = 10 + 5 = 15 \]

**First five terms**: 7, 9, 11, 13, 15

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### (ii) \( a_n = n(n + 2) \)

1. For \( n = 1 \):
\[ a_1 = 1(1 + 2) = 1 \cdot 3 = 3 \]

2. For \( n = 2 \):
\[ a_2 = 2(2 + 2) = 2 \cdot 4 = 8 \]

3. For \( n = 3 \):
\[ a_3 = 3(3 + 2) = 3 \cdot 5 = 15 \]

4. For \( n = 4 \):
\[ a_4 = 4(4 + 2) = 4 \cdot 6 = 24 \]

5. For \( n = 5 \):
\[ a_5 = 5(5 + 2) = 5 \cdot 7 = 35 \]

**First five terms**: 3, 8, 15, 24, 35

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### (iii) \( a_n = \frac{n - 3}{4} \)

1. For \( n = 1 \):
\[ a_1 = \frac{1 - 3}{4} = \frac{-2}{4} = -\frac{1}{2} \]

2. For \( n = 2 \):
\[ a_2 = \frac{2 - 3}{4} = \frac{-1}{4} \]

3. For \( n = 3 \):
\[ a_3 = \frac{3 - 3}{4} = \frac{0}{4} = 0 \]

4. For \( n = 4 \):
\[ a_4 = \frac{4 - 3}{4} = \frac{1}{4} \]

5. For \( n = 5 \):
\[ a_5 = \frac{5 - 3}{4} = \frac{2}{4} = \frac{1}{2} \]

**First five terms**: -1/2, -1/4, 0, 1/4, 1/2

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### (iv) \( a_n = \frac{n}{n + 1} \)

1. For \( n = 1 \):
\[ a_1 = \frac{1}{1 + 1} = \frac{1}{2} \]

2. For \( n = 2 \):
\[ a_2 = \frac{2}{2 + 1} = \frac{2}{3} \]

3. For \( n = 3 \):
\[ a_3 = \frac{3}{3 + 1} = \frac{3}{4} \]

4. For \( n = 4 \):
\[ a_4 = \frac{4}{4 + 1} = \frac{4}{5} \]

5. For \( n = 5 \):
\[ a_5 = \frac{5}{5 + 1} = \frac{5}{6} \]

**First five terms**: 1/2, 2/3, 3/4, 4/5, 5/6

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### (v) \( a_n = \frac{n(n^2 + 5)}{4} \)

1. For \( n = 1 \):
\[ a_1 = \frac{1(1^2 + 5)}{4} = \frac{1 \cdot 6}{4} = \frac{6}{4} = \frac{3}{2} \]

2. For \( n = 2 \):
\[ a_2 = \frac{2(2^2 + 5)}{4} = \frac{2 \cdot 9}{4} = \frac{18}{4} = \frac{9}{2} \]

3. For \( n = 3 \):
\[ a_3 = \frac{3(3^2 + 5)}{4} = \frac{3 \cdot 14}{4} = \frac{42}{4} = \frac{21}{2} \]

4. For \( n = 4 \):
\[ a_4 = \frac{4(4^2 + 5)}{4} = \frac{4 \cdot 21}{4} = 21 \]

5. For \( n = 5 \):
\[ a_5 = \frac{5(5^2 + 5)}{4} = \frac{5 \cdot 30}{4} = \frac{150}{4} = \frac{75}{2} \]

**First five terms**: 3/2, 9/2, 21/2, 21, 75/2

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