Question

пожалуйста помогите ​

Answer 1

Ответ:

49. A) 3

50. D) 65

51. D) 2/3

Объяснение:

49.

$b_2= \sqrt[7]{2^6}$

$b_3=2$

$q=\frac{b_3}{b_2}=\frac{2^1}{2^{\frac{6}{7}}}=2^{\frac{1}{7}}$

$S_{14}=\frac{b_1((2^{\frac{1}{7}})^{14}-1)}{2^{\frac{1}{7}}-1}=\frac{b_1(2^2-1)}{2^{\frac{1}{7}}-1}=\frac{b_1(4-1)}{2^{\frac{1}{7}}-1}=\frac{3b_1}{2^{\frac{1}{7}}-1}$

$S_7=\frac{b_1((2^{\frac{1}{7}})^{7}-1)}{2^{\frac{1}{7}}-1}=\frac{b_1(2-1)}{2^{\frac{1}{7}}-1}=\frac{b_1}{2^{\frac{1}{7}}-1}$

$\frac{S_{14}}{S_7}=\frac{\frac{3b_1}{2^{\frac{1}{7}}-1}}{\frac{b_1}{2^{\frac{1}{7}}-1}}=\frac{3b_1}{2^{\frac{1}{7}}-1}\cdot\frac{2^{\frac{1}{7}}-1}{b_1}=3$

50.

$b_2=3$

$b_3=6$

$q=\frac{b_3}{b_2}=\frac{6}{3}=2$

$S_{12}:S_6=\frac{b_1(2^{12}-1)}{2-1}:\frac{b_1(2^{6}-1)}{2-1}=\frac{b_1(2^{12}-1)}{1}:\frac{b_1(2^{6}-1)}{1}=\frac{b_1(2^{12}-1)}{b_1(2^{6}-1)}=\\\\ \frac{2^{12}-1}{2^{6}-1}=\frac{(2^{6}-1)(2^6+1)}{2^{6}-1}=2^6+1=64+1=65$

51.

$S_n=3-3^{1-n}$

$b_2=S_2-S_1=3-3^{1-2}-(3-3^{1-1})=3-3^{-1}-(3-3^{0})=3-\frac{1}{3}-(3-1)=2\frac{2}{3}-2=\frac{2}{3}$