To find the first four terms of the sequences defined by the given nth terms, we will substitute values of \( n \) from 1 to 4 into each expression.
### (i) \( 2^n \)
1. For \( n = 1 \):
\[
a_1 = 2^1 = 2
\]
2. For \( n = 2 \):
\[
a_2 = 2^2 = 4
\]
3. For \( n = 3 \):
\[
a_3 = 2^3 = 8
\]
4. For \( n = 4 \):
\[
a_4 = 2^4 = 16
\]
**First four terms**: \( 2, 4, 8, 16 \)
### (ii) \( \frac{n}{n+1} \)
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 \( n = 4 \):
\[
a_4 = \frac{4}{4+1} = \frac{4}{5}
\]
**First four terms**: \( \frac{1}{2}, \frac{2}{3}, \frac{3}{4}, \frac{4}{5} \)
### (iii) \( n^2 - 16 \)
1. For \( n = 1 \):
\[
a_1 = 1^2 - 16 = 1 - 16 = -15
\]
2. For \( n = 2 \):
\[
a_2 = 2^2 - 16 = 4 - 16 = -12
\]
3. For \( n = 3 \):
\[
a_3 = 3^2 - 16 = 9 - 16 = -7
\]
4. For \( n = 4 \):
\[
a_4 = 4^2 - 16 = 16 - 16 = 0
\]
**First four terms**: \( -15, -12, -7, 0 \)
### (iv) \( \frac{3^n}{2^n + 1} \)
1. For \( n = 1 \):
\[
a_1 = \frac{3^1}{2^1 + 1} = \frac{3}{2 + 1} = \frac{3}{3} = 1
\]
2. For \( n = 2 \):
\[
a_2 = \frac{3^2}{2^2 + 1} = \frac{9}{4 + 1} = \frac{9}{5}
\]
3. For \( n = 3 \):
\[
a_3 = \frac{3^3}{2^3 + 1} = \frac{27}{8 + 1} = \frac{27}{9} = 3
\]
4. For \( n = 4 \):
\[
a_4 = \frac{3^4}{2^4 + 1} = \frac{81}{16 + 1} = \frac{81}{17}
\]
**First four terms**: \( 1, \frac{9}{5}, 3, \frac{81}{17} \)
### (v) \( \frac{n+4}{n+1} \)
1. For \( n = 1 \):
\[
a_1 = \frac{1+4}{1+1} = \frac{5}{2}
\]
2. For \( n = 2 \):
\[
a_2 = \frac{2+4}{2+1} = \frac{6}{3} = 2
\]
3. For \( n = 3 \):
\[
a_3 = \frac{3+4}{3+1} = \frac{7}{4}
\]
4. For \( n = 4 \):
\[
a_4 = \frac{4+4}{4+1} = \frac{8}{5}
\]
**First four terms**: \( \frac{5}{2}, 2, \frac{7}{4}, \frac{8}{5} \)
### (vi) \( \log(1 + \frac{1}{n}) \)
1. For \( n = 1 \):
\[
a_1 = \log(1 + 1) = \log(2)
\]
2. For \( n = 2 \):
\[
a_2 = \log(1 + \frac{1}{2}) = \log\left(\frac{3}{2}\right)
\]
3. For \( n = 3 \):
\[
a_3 = \log(1 + \frac{1}{3}) = \log\left(\frac{4}{3}\right)
\]
4. For \( n = 4 \):
\[
a_4 = \log(1 + \frac{1}{4}) = \log\left(\frac{5}{4}\right)
\]
**First four terms**: \( \log(2), \log\left(\frac{3}{2}\right), \log\left(\frac{4}{3}\right), \log\left(\frac{5}{4}\right) \)
### Summary of First Four Terms
1. \( 2, 4, 8, 16 \)
2. \( \frac{1}{2}, \frac{2}{3}, \frac{3}{4}, \frac{4}{5} \)
3. \( -15, -12, -7, 0 \)
4. \( 1, \frac{9}{5}, 3, \frac{81}{17} \)
5. \( \frac{5}{2}, 2, \frac{7}{4}, \frac{8}{5} \)
6. \( \log(2), \log\left(\frac{3}{2}\right), \log\left(\frac{4}{3}\right), \log\left(\frac{5}{4}\right) \)
