Задача 69662 Не могу решить ...
Условие
Решение
[m]S(5) = b1 \cdot \frac{q^5 - 1}{q-1} = 2\sqrt{3} \cdot \frac{\sqrt{3}^5 - 1}{\sqrt{3}-1} = 2\sqrt{3} \cdot \frac{(9\sqrt{3} - 1)(\sqrt{3}+1)}{(\sqrt{3}-1)(\sqrt{3}+1)} = [/m]
[m] = 2\sqrt{3} \cdot \frac{9\cdot 3 - \sqrt{3}+9\sqrt{3}-1)}{3-1}= 2\sqrt{3} \cdot \frac{26 + 8\sqrt{3}}{2} =[/m]
[m]= 2\sqrt{3}(13+4\sqrt{3}) = 26\sqrt{3}+8 \cdot 3 = 24+26\sqrt{3}[/m]
2) b1 = 5; q = -2,5
[m]S(6) = b1 \cdot \frac{q^6 - 1}{q-1} = 5 \cdot \frac{(-2,5)^6 - 1}{-2,5-1} = 5 \cdot \frac{243,14}{-3,5} = -5 \cdot 69,47 = -347,34[/m]
3) a1 = 3; q = -2
[m]S(4) = a1 \cdot \frac{q^4 - 1}{q-1} = 3 \cdot \frac{(-2)^4 - 1}{-2-1} = 3 \cdot \frac{16 - 1}{-3} = 3\cdot (-5) = -15[/m]
4) q = 2/3; S(4) = -65
[m]S(4) = c1 \cdot \frac{q^4 - 1}{q-1}[/m]
[m]-65 = c1 \cdot \frac{(2/3)^4 - 1}{2/3-1} = c1 \cdot \frac{16/81 - 1}{-1/3} = c1 \cdot \frac{16-81}{81} \cdot (-3) = c1 \cdot \frac{-65}{-27}[/m]
[m]c1 = \frac{(-65)(-27)}{-65} = -27[/m]