670. Вычислите первые пять членов последовательности (с), заданной формулой: a) cn = -2n² + 7; 6) Cn = 100 n²-5'; B) Cn = -2,5 * 2"; г) сп = 3,2. 2- "; д) Сп = (-1)-1 -; 4n e) Cn = 1 – (−1)" 2n + 1.

a) \( c_n = -2n^2 + 7 \)

1. \( c_1 = -2(1)^2 + 7 = -2 + 7 = 5 \)
2. \( c_2 = -2(2)^2 + 7 = -8 + 7 = -1 \)
3. \( c_3 = -2(3)^2 + 7 = -18 + 7 = -11 \)
4. \( c_4 = -2(4)^2 + 7 = -32 + 7 = -25 \)
5. \( c_5 = -2(5)^2 + 7 = -50 + 7 = -43 \)

б) \( c_n = \frac{100}{n^2 - 5} \)

1. \( c_1 = \frac{100}{1^2 - 5} = \frac{100}{-4} = -25 \)
2. \( c_2 = \frac{100}{2^2 - 5} = \frac{100}{-1} = -100 \)
3. \( c_3 = \frac{100}{3^2 - 5} = \frac{100}{4} = 25 \)
4. \( c_4 = \frac{100}{4^2 - 5} = \frac{100}{11} \)
5. \( c_5 = \frac{100}{5^2 - 5} = \frac{100}{20} = 5 \)

в) \( c_n = -2.5 \cdot 2^n \)

1. \( c_1 = -2.5 \cdot 2^1 = -5 \)
2. \( c_2 = -2.5 \cdot 2^2 = -10 \)
3. \( c_3 = -2.5 \cdot 2^3 = -20 \)
4. \( c_4 = -2.5 \cdot 2^4 = -40 \)
5. \( c_5 = -2.5 \cdot 2^5 = -80 \)

г) \( c_n = 3.2 \cdot 2^{-n} \)

1. \( c_1 = 3.2 \cdot 2^{-1} = 3.2 \cdot \frac{1}{2} = 1.6 \)
2. \( c_2 = 3.2 \cdot 2^{-2} = 3.2 \cdot \frac{1}{4} = 0.8 \)
3. \( c_3 = 3.2 \cdot 2^{-3} = 3.2 \cdot \frac{1}{8} = 0.4 \)
4. \( c_4 = 3.2 \cdot 2^{-4} = 3.2 \cdot \frac{1}{16} = 0.2 \)
5. \( c_5 = 3.2 \cdot 2^{-5} = 3.2 \cdot \frac{1}{32} = 0.1 \)

д) \( c_n = \frac{(-1)^{n-1}}{4n} \)

1. \( c_1 = \frac{(-1)^{1-1}}{4 \cdot 1} = \frac{1}{4} \)
2. \( c_2 = \frac{(-1)^{2-1}}{4 \cdot 2} = \frac{-1}{8} = -\frac{1}{8} \)
3. \( c_3 = \frac{(-1)^{3-1}}{4 \cdot 3} = \frac{1}{12} \)
4. \( c_4 = \frac{(-1)^{4-1}}{4 \cdot 4} = \frac{-1}{16} = -\frac{1}{16} \)
5. \( c_5 = \frac{(-1)^{5-1}}{4 \cdot 5} = \frac{1}{20} \)

e) \( c_n = \frac{1 - (-1)^n}{2n + 1} \)

1. \( c_1 = \frac{1 - (-1)^1}{2 \cdot 1 + 1} = \frac{1 + 1}{3} = \frac{2}{3} \)
2. \( c_2 = \frac{1 - (-1)^2}{2 \cdot 2 + 1} = \frac{1 - 1}{5} = 0 \)
3. \( c_3 = \frac{1 - (-1)^3}{2 \cdot 3 + 1} = \frac{1 + 1}{7} = \frac{2}{7} \)
4. \( c_4 = \frac{1 - (-1)^4}{2 \cdot 4 + 1} = \frac{1 - 1}{9} = 0 \)
5. \( c_5 = \frac{1 - (-1)^5}{2 \cdot 5 + 1} = \frac{1 + 1}{11} = \frac{2}{11} \)