Griffiths Barber
06/19/2023 · Senior High School
Ejercitación Encuentra los primeros cinco términos de cada su- \( \begin{array}{ll}\text { a. } a_{n}=3 n-2 & \text { b. } a_{n}=\frac{1}{3 n} \\ \text { c. } a_{n}=\frac{1}{n+1} & \text { d. } a_{n}=n^{2}+2 \\ \text { e. } a_{n}=\frac{2}{2 n+1} & \text { f. } a_{n}=\frac{2 n^{2}}{3}\end{array} \)
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Para encontrar los primeros cinco términos de cada una de las sucesiones dadas, simplemente sustituimos \( n \) por los valores 1, 2, 3, 4 y 5 en cada fórmula.
### a. \( a_{n} = 3n - 2 \)
- \( a_1 = 3(1) - 2 = 1 \)
- \( a_2 = 3(2) - 2 = 4 \)
- \( a_3 = 3(3) - 2 = 7 \)
- \( a_4 = 3(4) - 2 = 10 \)
- \( a_5 = 3(5) - 2 = 13 \)
**Primeros cinco términos:** 1, 4, 7, 10, 13
### b. \( a_{n} = \frac{1}{3n} \)
- \( a_1 = \frac{1}{3(1)} = \frac{1}{3} \)
- \( a_2 = \frac{1}{3(2)} = \frac{1}{6} \)
- \( a_3 = \frac{1}{3(3)} = \frac{1}{9} \)
- \( a_4 = \frac{1}{3(4)} = \frac{1}{12} \)
- \( a_5 = \frac{1}{3(5)} = \frac{1}{15} \)
**Primeros cinco términos:** \( \frac{1}{3}, \frac{1}{6}, \frac{1}{9}, \frac{1}{12}, \frac{1}{15} \)
### c. \( a_{n} = \frac{1}{n+1} \)
- \( a_1 = \frac{1}{1+1} = \frac{1}{2} \)
- \( a_2 = \frac{1}{2+1} = \frac{1}{3} \)
- \( a_3 = \frac{1}{3+1} = \frac{1}{4} \)
- \( a_4 = \frac{1}{4+1} = \frac{1}{5} \)
- \( a_5 = \frac{1}{5+1} = \frac{1}{6} \)
**Primeros cinco términos:** \( \frac{1}{2}, \frac{1}{3}, \frac{1}{4}, \frac{1}{5}, \frac{1}{6} \)
### d. \( a_{n} = n^{2} + 2 \)
- \( a_1 = 1^{2} + 2 = 3 \)
- \( a_2 = 2^{2} + 2 = 6 \)
- \( a_3 = 3^{2} + 2 = 11 \)
- \( a_4 = 4^{2} + 2 = 18 \)
- \( a_5 = 5^{2} + 2 = 27 \)
**Primeros cinco términos:** 3, 6, 11, 18, 27
### e. \( a_{n} = \frac{2}{2n+1} \)
- \( a_1 = \frac{2}{2(1)+1} = \frac{2}{3} \)
- \( a_2 = \frac{2}{2(2)+1} = \frac{2}{5} \)
- \( a_3 = \frac{2}{2(3)+1} = \frac{2}{7} \)
- \( a_4 = \frac{2}{2(4)+1} = \frac{2}{9} \)
- \( a_5 = \frac{2}{2(5)+1} = \frac{2}{11} \)
**Primeros cinco términos:** \( \frac{2}{3}, \frac{2}{5}, \frac{2}{7}, \frac{2}{9}, \frac{2}{11} \)
### f. \( a_{n} = \frac{2n^{2}}{3} \)
- \( a_1 = \frac{2(1)^{2}}{3} = \frac{2}{3} \)
- \( a_2 = \frac{2(2)^{2}}{3} = \frac{8}{3} \)
- \( a_3 = \frac{2(3)^{2}}{3} = 6 \)
- \( a_4 = \frac{
Quick Answer
### a. \( a_{n} = 3n - 2 \)
- Primeros cinco términos: 1, 4, 7, 10, 13
### b. \( a_{n} = \frac{1}{3n} \)
- Primeros cinco términos: \( \frac{1}{3}, \frac{1}{6}, \frac{1}{9}, \frac{1}{12}, \frac{1}{15} \)
### c. \( a_{n} = \frac{1}{n+1} \)
- Primeros cinco términos: \( \frac{1}{2}, \frac{1}{3}, \frac{1}{4}, \frac{1}{5}, \frac{1}{6} \)
### d. \( a_{n} = n^{2} + 2 \)
- Primeros cinco términos: 3, 6, 11, 18, 27
### e. \( a_{n} = \frac{2}{2n+1} \)
- Primeros cinco términos: \( \frac{2}{3}, \frac{2}{5}, \frac{2}{7}, \frac{2}{9}, \frac{2}{11} \)
### f. \( a_{n} = \frac{2n^{2}}{3} \)
- Primeros cinco términos: \( \frac{2}{3}, \frac{8}{3}, 6, \frac{16}{3}, \frac{24}{3} \)
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