Question

Halla los primeros 5 términos de cada sucesión:
1. an = n^2
2. an = n+5
3. an = n^2-2
4. bn = -2n-3/n+4
5. an = n^2/n+5
Gracias.

Answer 1

Hola!

Halla los primeros 5 términos de cada sucesión:

1. an = n^2

2. an = n+5

3. an = n^2-2

4. bn = -2n-3/n+4

5. an = n^2/n+5

Solución:

• 1. an = n^2

$a_n = n^2$

$a_1 = 1^2 \to \boxed{a_1 = 1}$

$a_2 = 2^2 \to \boxed{a_2 = 4}$

$a_3 = 3^2 \to \boxed{a_3 = 9}$

$a_4 = 4^2 \to \boxed{a_4 = 16}$

$a_5 = 5^2 \to \boxed{a_5 = 25}$

• 2. an = n+5

$a_n = n+5$

$a_1 = 1+5 \to \boxed{a_1 = 6}$

$a_2 = 2+5 \to \boxed{a_2 = 7}$

$a_3 = 3+5 \to \boxed{a_3 = 8}$

$a_4 = 4+5 \to \boxed{a_4 = 9}$

$a_5 = 5+5 \to \boxed{a_5 = 10}$

• 3. an = n^2-2

$a_n = n^2 - 2$

$a_1 = 1^2 - 2 \to a_1 = 1 - 2 \to \boxed{a_1 = -1}$

$a_2 = 2^2 - 2 \to a_2 = 4 - 2 \to \boxed{a_2 = 2}$

$a_3 = 3^2 - 2 \to a_3 = 9 - 2 \to \boxed{a_3 = 7}$

$a_4 = 4^2 - 2 \to a_4 = 16 - 2 \to \boxed{a_4 = 14}$

$a_5 = 5^2 - 2 \to a_5 = 25 - 2 \to \boxed{a_5 = 23}$

• 4. bn = -2n-3/n+4

$b_n = \dfrac{-2n-3}{n+4}$

$b_1 = \dfrac{-2*1-3}{1+4} \to b_1 = \dfrac{-2-3}{5} \to b_1 = \dfrac{-5}{5} \to \boxed{b_1 = -1}$

$b_2 = \dfrac{-2*2-3}{2+4} \to b_2 = \dfrac{-4-3}{6} \to \boxed{b_2 = \dfrac{-7}{6}}$

$b_3 = \dfrac{-2*3-3}{3+4} \to b_3 = \dfrac{-6-3}{7} \to \boxed{b_3 = \dfrac{-9}{7}}$

$b_4 = \dfrac{-2*4-3}{4+4} \to b_4 = \dfrac{-8-3}{8} \to \boxed{b_4 = \dfrac{-11}{8}}$

$b_5 = \dfrac{-2*5-3}{5+4} \to b_5 = \dfrac{-10-3}{9} \to \boxed{b_5 = \dfrac{-13}{9}}$

• 5. an = n^2/n+5

$a_n = \dfrac{n^2}{n+5}$

$a_1 = \dfrac{1^2}{1+5} \to \boxed{a_1 = \dfrac{1}{6}}$

$a_2 = \dfrac{2^2}{2+5} \to \boxed{a_2 = \dfrac{4}{7}}$

$a_3 = \dfrac{3^2}{3+5} \to \boxed{a_3 = \dfrac{9}{8}}$

$a_4 = \dfrac{4^2}{4+5} \to \boxed{a_4 = \dfrac{16}{9}}$

$a_5 = \dfrac{5^2}{5+5} \to a_5 = \dfrac{25}{10}\frac{\div5}{\div5} \to \boxed{a_5 = \dfrac{5}{2}}$

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Answer 2

Respuesta:

1. an = n^2

2. an = n+5

3. an = n^2-2

4. bn = -2n-3/n+4

5. an = n^2/n+5

Explicación paso a paso: