Para hallar los primeros cinco términos, evaluamos el término general $a_n$ para $n=1, 2, 3, 4, 5$.

(a) $a_n = 1 + \frac{1}{n}$

• $n=1: 1+1=2$
• $n=2: 1+\frac{1}{2}=\frac{3}{2}$
• $n=3: 1+\frac{1}{3}=\frac{4}{3}$
• $n=4: 1+\frac{1}{4}=\frac{5}{4}$
• $n=5: 1+\frac{1}{5}=\frac{6}{5}$

$$ \boxed{2, \frac{3}{2}, \frac{4}{3}, \frac{5}{4}, \frac{6}{5}} $$

(b) $a_n = \frac{1}{n(n+1)}$

• $n=1: \frac{1}{1(2)}=\frac{1}{2}$
• $n=2: \frac{1}{2(3)}=\frac{1}{6}$
• $n=3: \frac{1}{3(4)}=\frac{1}{12}$
• $n=4: \frac{1}{4(5)}=\frac{1}{20}$
• $n=5: \frac{1}{5(6)}=\frac{1}{30}$

$$ \boxed{\frac{1}{2}, \frac{1}{6}, \frac{1}{12}, \frac{1}{20}, \frac{1}{30}} $$

(c) $a_n = a + (n-1)d$ (Progresión Aritmética)

• $n=1: a+0d = a$
• $n=2: a+1d$
• $n=3: a+2d$
• $n=4: a+3d$
• $n=5: a+4d$

$$ \boxed{a, a+d, a+2d, a+3d, a+4d} $$

(d) $a_n = (-1)^{n+1} ar^{n-1}$ (Progresión Geométrica Alternada)

• $n=1: (-1)^2 ar^0 = a$
• $n=2: (-1)^3 ar^1 = -ar$
• $n=3: (-1)^4 ar^2 = ar^2$
• $n=4: (-1)^5 ar^3 = -ar^3$
• $n=5: (-1)^6 ar^4 = ar^4$

$$ \boxed{a, -ar, ar^2, -ar^3, ar^4} $$

(e) $a_n = \frac{n}{\sqrt{1+n^2}}$

• $n=1: \frac{1}{\sqrt{1+1}} = \frac{1}{\sqrt{2}}$
• $n=2: \frac{2}{\sqrt{1+4}} = \frac{2}{\sqrt{5}}$
• $n=3: \frac{3}{\sqrt{1+9}} = \frac{3}{\sqrt{10}}$
• $n=4: \frac{4}{\sqrt{1+16}} = \frac{4}{\sqrt{17}}$
• $n=5: \frac{5}{\sqrt{1+25}} = \frac{5}{\sqrt{26}}$

$$ \boxed{\frac{1}{\sqrt{2}}, \frac{2}{\sqrt{5}}, \frac{3}{\sqrt{10}}, \frac{4}{\sqrt{17}}, \frac{5}{\sqrt{26}}} $$

(f) $a_n = \frac{\sqrt{n+1}}{n}$

• $n=1: \frac{\sqrt{2}}{1} = \sqrt{2}$
• $n=2: \frac{\sqrt{3}}{2}$
• $n=3: \frac{\sqrt{4}}{3} = \frac{2}{3}$
• $n=4: \frac{\sqrt{5}}{4}$
• $n=5: \frac{\sqrt{6}}{5}$

$$ \boxed{\sqrt{2}, \frac{\sqrt{3}}{2}, \frac{2}{3}, \frac{\sqrt{5}}{4}, \frac{\sqrt{6}}{5}} $$

(g) $a_n = (-1)^{n+1} \frac{n!}{n^n}$

• $n=1: (-1)^2 \frac{1!}{1^1} = 1$
• $n=2: (-1)^3 \frac{2!}{2^2} = -\frac{2}{4} = -\frac{1}{2}$
• $n=3: (-1)^4 \frac{3!}{3^3} = \frac{6}{27} = \frac{2}{9}$
• $n=4: (-1)^5 \frac{4!}{4^4} = -\frac{24}{256} = -\frac{3}{32}$
• $n=5: (-1)^6 \frac{5!}{5^5} = \frac{120}{3125} = \frac{24}{625}$

$$ \boxed{1, -\frac{1}{2}, \frac{2}{9}, -\frac{3}{32}, \frac{24}{625}} $$

(h) $a_n = \frac{(2n)!}{3^n 5^{n-1}}$

• $n=1: \frac{2!}{3^1 5^0} = \frac{2}{3}$
• $n=2: \frac{4!}{3^2 5^1} = \frac{24}{45} = \frac{8}{15}$ (u expresado como $\frac{2^3}{3 \cdot 5}$)
• $n=3: \frac{6!}{3^3 5^2} = \frac{720}{27 \cdot 25} = \frac{720}{675} = \frac{16}{15}$ (u expresado como $\frac{2^4}{3 \cdot 5}$)
• $n=4: \frac{8!}{3^4 5^3} = \frac{40320}{81 \cdot 125} = \frac{40320}{10125} = \frac{896}{225} = \frac{7 \cdot 2^7}{3^2 \cdot 5^2}$
• $n=5: \frac{10!}{3^5 5^4} = \frac{3628800}{243 \cdot 625} = \frac{3628800}{151875} = \frac{1792}{75} = \frac{7 \cdot 2^8}{3 \cdot 5^2}$

$$ \boxed{\frac{2}{3}, \frac{2^3}{3 \cdot 5}, \frac{2^4}{3 \cdot 5}, \frac{7 \cdot 2^7}{3^2 \cdot 5^2}, \frac{7 \cdot 2^8}{3 \cdot 5^2}} $$