Найти сумму пяти первых членов прогрессии (bn), у которой b1 = 3, а q = 2.

1)  27*2^x-8*3^x=0  /3^x
27*(2/3)^x - 8 = 0
(2/3)^x = 8/27
(2/3)^x = (2/3)^3
x = 3
ответ: х = 3

2)  2^(x+1) - 2^(x-1)=3^(2-x)
2*(2^x) - (1/2)*(2^x)  = 9/(3^x)
(2^x) *(2 - 1/2) = 9/(3^x)
(2^x)*(3/2) = 9/(3/2)   
(6^x) = 6^1
x = 1
ответ: х = 1

3)  9*(4^x)  - 13*(6^x) + 4*(9^x) = 0
 9*(2^2x)  - 13*(2^x)*(3^x) + 4*(3^2x) = 0   /(3^2x)
9*(2/3)^2x - 13*(2/3)^x + 4 = 0
(2/3)^x = t
9t^2 - 13t + 4 = 0
D = 169 - 4*9*4 = 25
t1 = (13 - 5)/18
t1 = 4/9
t2 = (13 + 5)/18 
t2 = 1
1)  (2/3)^x = 4/9
(2/3)^x = (2/3)^2
x1 = 2
2)  (2/3)^x = 1
(2/3)^x = (2/3)^0
x2 = 0
ответ: x1 = 2; x2 = 1

1)  27*2^x-8*3^x=0  /3^x
27*(2/3)^x - 8 = 0
(2/3)^x = 8/27
(2/3)^x = (2/3)^3
x = 3
ответ: х = 3

2)  2^(x+1) - 2^(x-1)=3^(2-x)
2*(2^x) - (1/2)*(2^x)  = 9/(3^x)
(2^x) *(2 - 1/2) = 9/(3^x)
(2^x)*(3/2) = 9/(3/2)   
(6^x) = 6^1
x = 1
ответ: х = 1

3)  9*(4^x)  - 13*(6^x) + 4*(9^x) = 0
 9*(2^2x)  - 13*(2^x)*(3^x) + 4*(3^2x) = 0   /(3^2x)
9*(2/3)^2x - 13*(2/3)^x + 4 = 0
(2/3)^x = t
9t^2 - 13t + 4 = 0
D = 169 - 4*9*4 = 25
t1 = (13 - 5)/18
t1 = 4/9
t2 = (13 + 5)/18 
t2 = 1
1)  (2/3)^x = 4/9
(2/3)^x = (2/3)^2
x1 = 2
2)  (2/3)^x = 1
(2/3)^x = (2/3)^0
x2 = 0
ответ: x1 = 2; x2 = 1