Алгебра 9 класс учебник Макарычев, Миндюк ответы – номер 677

а) q = –1/3, n = 5, S_n = 201/3; найдите х₁ и х_n;

S_n = $$\frac{x_1(q^n - 1)}{q - 1}$$

201/3 = $$\frac{x_1((-\frac{1}{3})^{5} - 1)}{-\frac{1}{3} - 1}$$

61/3 = $$\frac{x_1(-\frac{1}{243} - 1)}{-\frac{4}{3}}$$

х₁ · (–244/243) = 61/3 · (–4/3)

х₁ · (–244/243) = –244/9

х₁ = –244/9 : (–244/243) = 244/9 · 244/243 = 243/9 = 27

х_n = х₁q^{n - 1}

х₅ = 27 · (–1/3)⁴ = 27 · 1/81 = 1/3

Ответ: х₁ = 27; х₅ = 1/3

б) х₁ = 11, х_n = 88, S_n = 165

S_n = $$\frac{x_nq - 1_1}{q - 1}$$

165 = 88q - 11/q - 1

165(q – 1) = 88q – 11
165q – 88q = –11 + 165
77q = 154
q = 2
х_n = х₁q^{n - 1}
88 = 11 · 2^{n - 1}
2^{n - 1} = 88 : 11
2^{n - 1} = 8
2^{n - 1} = 2³
n – 1 = 3
n = 4

Ответ: q = 2; n = 4

в) х₁ = 1/2, q = –1/2, S_n = 21/64
S_n = $$\frac{x_1(q^n - 1)}{q - 1}$$

21/64 = $$\frac{\frac{1}{2}((-\frac{1}{2})^{n} - 1)}{-\frac{1}{2} - 1}$$

21/64 = $$\frac{\frac{1}{2}((-\frac{1}{2})^{n} - 1)}{-\frac{3}{2}}$$

21/64 = $$\frac{1 · (\frac{1}{2})^{n} - 1}{-3}$$

(1/2)_n – 1 = 21 · (–3) : 64
(1/2)_n – 1 = –63/64
(1/2)_n = 1 – 63/64
(1/2)_n = 1/64
(1/2)_n = (1/2)₆
n = 6
х_n = x₁q^{n - 1}
х_n = x₁q^{n - 1}
х₆ = 1/2 · (–1/6)⁵
x₆ = –(1/2)⁶
x₆ = –1/64

Ответ: n = 6; х₆ = –1/64

г) q = $$\sqrt{3}$$, х_n = 18$$\sqrt{3}$$, S_n = 26$$\sqrt{3}$$ + 24

S_n = $$\frac{x_{nq} - x_1}{q - 1}$$

26$$\sqrt{3}$$ + 24 = $$\frac{18\sqrt{3} · \sqrt{3} - x_1}{\sqrt{3} - 1}$$

26$$\sqrt{3}$$ + 24 = $$\frac{54 - x_1}{\sqrt{3} - 1}$$

54 – x₁ = (26$$\sqrt{3}$$ + 24)($$\sqrt{3}$$ – 1)

54 – x₁ = 26 · $$\sqrt{3}$$ + $$\sqrt{3}$$ + 24$$\sqrt{3}$$ – 26$$\sqrt{3}$$ – 24

54 – x₁ = 78 – 2$$\sqrt{3}$$ – 24

54 – x₁ = 54 – 2$$\sqrt{3}$$

x₁ = 2$$\sqrt{3}$$

x_n = x₁q^{n - 1}

18$$\sqrt{3}$$ = 2$$\sqrt{3}$$ · ($$\sqrt{3}$$)^{n - 1}

($$\sqrt{3}$$)^{n - 1} = 9

(31/2)^{n - 1} = 3²



0,5n – 0,5 = 2
0,5n = 2,5
n = 5

Ответ: x₁ = 2$$\sqrt{3}$$; n = 5