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標題:
geometric sequence
發問:
1. Given a geometric sequence whose sum of the first 10 terms is 4 and whose sum from the 11th to the 30th term is 48, find the sum from the 31st to the 60th term.
最佳解答:
4 = a(r^10 - 1)/(r - 1) , 48+4 = 52 = a(r^30 - 1)/(r-1) so 13 = (r^30 - 1)/(r^10 -1) r^10 = 1(rej), 3, -4(rej) Therefore sum = a(r^60 -1)/(r-1) - 48 = a(r^30 + 1)(r^30 - 1)/(r-1) - 48 = 52r^30+4 = 472
其他解答:
geometric sequence
發問:
1. Given a geometric sequence whose sum of the first 10 terms is 4 and whose sum from the 11th to the 30th term is 48, find the sum from the 31st to the 60th term.
最佳解答:
4 = a(r^10 - 1)/(r - 1) , 48+4 = 52 = a(r^30 - 1)/(r-1) so 13 = (r^30 - 1)/(r^10 -1) r^10 = 1(rej), 3, -4(rej) Therefore sum = a(r^60 -1)/(r-1) - 48 = a(r^30 + 1)(r^30 - 1)/(r-1) - 48 = 52r^30+4 = 472
其他解答:
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