Find the sum of the geometric series for which a1 = 125, r = ⅖, and n
= 4.?
回答 (4)
S(n) = a₁ [1 - rⁿ] / [1 - r]
S(4)
= 125 × [1 - (2/5)⁴] / [1 - (2/5)]
= 125 × (609/625) / (3/5)
= 203
sum of geometric series
= a1 * (1 - r^n) / (1 - r)
= (125) * (1 - (2/5)^4) / (1 - (2/5))
= (125) * (1 - (16/625)) / (3/5)
= (125 - (16/5)) / (3/5)
= (625 - 16) / 3
= 609 / 3
= 203
收錄日期: 2021-04-18 14:58:19
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20160609190008AAk4PxN
檢視 Wayback Machine 備份