Find the sum of the geometric series for which a1 = 125, r = ⅖, and n = 4.?
回答 (4)
2016-06-10 3:07 am
S(n) = a₁ [1 - rⁿ] / [1 - r]
S(4)
= 125 × [1 - (2/5)⁴] / [1 - (2/5)]
= 125 × (609/625) / (3/5)
= 203
S(4)
= 125 × [1 - (2/5)⁴] / [1 - (2/5)]
= 125 × (609/625) / (3/5)
= 203
2016-06-10 3:07 am
203
2016-06-10 3:01 am
69
2016-06-10 3:52 am
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
= 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
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