Упр.17.30 ГДЗ Мордкович Александрова 9 класс (Алгебра)

1. Для геометрической прогрессии используем формулы

   $$b_n=b_1q^{n-1}, \qquad S_n=\frac{b_1(q^n-1)}{q-1}\ (q\ne 1).$$

   Тогда получаем:

   $$b_1$$	$$q$$	$$n$$	$$b_n$$	$$S_n$$
   $$15$$	$$\frac13$$	$$3$$	$$\frac53$$	$$21\frac23$$
   $$2$$	$$3$$	$$3$$	$$18$$	$$26$$
   $$\frac13$$	$$1\frac12$$	$$6$$	$$2\frac{17}{32}$$	$$\frac{121}{24}$$
   $$\sqrt3$$	$$\sqrt3$$	$$4$$	$$9$$	$$4(3+\sqrt3)$$
   $$15$$	$$\frac13$$	$$6$$	$$\frac{5}{81}$$	$$\frac{605}{27}$$
   $$\frac{25}{169}$$	$$\frac{13}{5}$$	$$4$$	$$\frac{13}{5}$$	$$\frac{3492}{169}$$
   $$2\sqrt6$$	$$\frac{1}{\sqrt6}$$	$$4$$	$$\frac13$$	$$\frac{7(\sqrt6+1)}{3}$$

Ответ

$$\begin{array}{c|c|c|c|c} b_1 & q & n & b_n & S_n\\ \hline 15 & \frac13 & 3 & \frac53 & 21\frac23\\ 2 & 3 & 3 & 18 & 26\\ \frac13 & 1\frac12 & 6 & 2\frac{17}{32} & \frac{121}{24}\\ \sqrt3 & \sqrt3 & 4 & 9 & 4(3+\sqrt3)\\ 15 & \frac13 & 6 & \frac{5}{81} & \frac{605}{27}\\ \frac{25}{169} & \frac{13}{5} & 4 & \frac{13}{5} & \frac{3492}{169}\\ 2\sqrt6 & \frac{1}{\sqrt6} & 4 & \frac13 & \frac{7(\sqrt6+1)}{3} \end{array}$$