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wanwai 2017-6-27 10:05 PM

G.S等比數列

[attach]7020942[/attach]
各位師兄,請教一下
諗爆頭了

ppresent 2017-6-27 11:26 PM

G.S. 1+x+x^2+x^3+.....=1/(1-x) for abs(x)<1

vertical distance travelled
=3.5(2)+3.5(0.75)(2)+3.5(0.75^2)(2)+... (moves up&down, dist=height*2)
=7(1+0.75+0.75^2+...)
=?

$holder 2017-7-21 02:31 AM

1:可視為無限項
2:use formula: a/1-r
(r=.75)
Ans:14m

啱5啱?

自己諗
...DSE得4
麻煩樓主有答案時通知下我:smile_o06:

[[i] 本帖最後由 $holder 於 2017-7-21 02:34 AM 編輯 [/i]]

Jonleefun 2017-7-23 06:43 AM

let's consider downward travel
=3.5+3.5*0.75+3.5*0.75*0.75+.....)
=3.5(1+0.75+0.75*0.75+.....)
=3.5(1+0.75+0.75^2+.....)
by using (1-a^n)=(1-a)(1+a+a^2+...)
(1+a+a^2+...)=(1-a^n)/(1-a)
and if n -> infinite, and a<1
then
(1+a+a^2+...)=1/(1-a)
Therefore
=3.5(1+0.75+0.75^2+.....)
=3.5(1)/(1-0.75)
=3.5/0.25
=14m

downward travel = upward travel
Total travel = downward travel + upward travel
the total travel = 28m

[[i] 本帖最後由 Jonleefun 於 2017-7-23 08:38 AM 編輯 [/i]]

$holder 2017-7-24 12:06 AM

難怪總覺得好似差咁D咩.....原本我只計了一半...:smile_o03:


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