數學知識交流---等比數列和(2)

(1) 求 5 - 15 + 45 - 135 + 405 -1215 + .. - 5 × 3^51。

等比數列和 = a(r^n - 1)/(r - 1)

5 - 15 + 45 - 135 + 405 - 1215 + ... - 5 × 3^51
= 5 - 15 + 45 - 135 + 405 - 1215 + ... + 5 x 3^50 - 5 × 3^51
= (5 + 45 + 405 + ... + 5 x 3^50) - (15 + 135 + 1215 ... + 5 x 3^51)
= (5 + 5 x 9 + 5 x 9^2 + ... 5 x 9^25) - (15 + 15 x 9 + 15 x 9^2 + ... 15 x9^25)
= [5 x (9^26 - 1)/(9 - 1)] - [15(9^26 - 1)/(9 - 1)]
= -10(9^26 - 1)/8
= -8076352361533341623665300


= = = = =
(2) 求(3^1)/(7^2) + (3^2)/(7^3) + (3^3)/(7^4) + (3^4)/(7^5) +...

等比數列的無限項和 = a/(1 - r)
首項 a = 3/49, 公比 r = 3/7

(3^1)/(7^2) + (3^2)/(7^3) + (3^3)/(7^4) + (3^4)/(7^5) + ...
= (3/49) / [1 - (3/7)]
= (3/49) / (4/7)
= (3/49) x (7/4)
= 3/28

2011-06-17 20:28:57 補充：
第 (2) 題另一方法：

設 S = (3^1)/(7^2) + (3^2)/(7^3) + (3^3)/(7^4) + (3^4)/(7^5) +...
(3/7)S = (3^2)/(7^3) + (3^3)/(7^4) + (3^4)/(7^5) +...

兩式相減：
(4/7)S = 3/49
S = 3/28 ..... 答案