In the formula of summation of GS,
Sum of the series is given by S(n) :
S(n) = a(r^n-1) / (r-1)
what if r=0,
i.e G.S.= a, a, a, a, a,….
By common sense, S(n) = na
But we put r=1, the formula is invalid, why?
in the formular of summation of GS
2007-10-11 7:04 am
回答 (2)
2007-10-11 7:20 am
✔ 最佳答案
You are wrong that when r = 0, G.S. = a,0,0,0,.......S(∞) = a
When r = 1, S(n) = na
2007-10-11 7:16 am
用你個概念諗
first term = a
common difference = 0
a, ar, ar^2, ...
a 0, 0, ...
唔係 a, a, a, ...
-------------------------
而且我地都唔會當 a 0, 0, ... 係 G.P.
公比 = 第二項/第一項 = 第三項/第二項
第三項/第二項 = 0/0
零除零, 你話咩黎呀?
明未??
唔好諗左 d 牛角尖,
比 d 心機諗數書 d 問題仲好
first term = a
common difference = 0
a, ar, ar^2, ...
a 0, 0, ...
唔係 a, a, a, ...
-------------------------
而且我地都唔會當 a 0, 0, ... 係 G.P.
公比 = 第二項/第一項 = 第三項/第二項
第三項/第二項 = 0/0
零除零, 你話咩黎呀?
明未??
唔好諗左 d 牛角尖,
比 d 心機諗數書 d 問題仲好
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