微積分問題 find sum of series

👨🏻‍🏫 學術台數學

[匿名]

2009-02-10 11:27 pm

using term by term integration for an appropriate series find the sum of the series

圖: http://www.geocities.com/tweety90331/01.GIF
(hint: factor first x to see "an appropriate series")

更新1:

其他兩個待解問題能請高手幫忙解答嗎非常感謝 http://hk.knowledge.yahoo.com/question/question?qid=7009021000749 http://hk.knowledge.yahoo.com/question/question?qid=7009021000756

回答 (2)

mathmanliu

2009-02-10 11:35 pm

✔ 最佳答案

1+x+x^2+.. = 1/(1-x) for -1<x<1
differtiating it, then 1+2x+ 3x^2 + ... = 1/(1-x)^2
=> x(1+2x+3x^2+ ...) = x/(1-x)^2
so, x+ 2x^2 + 3x^3+ ... = x/(1-x)^2

👍 0 👎 0

Ahliu

2009-02-10 11:39 pm

(assumed |x|<1)

Firstly, 1/ (1-x) = sum x^n from n=0 to infinity

now
x + 2x^2 + 3x^3 + ... + nx^n + ...
= x (2x + 3x^2 + nx^(n-1) + .... )
= x [sum nx^(n-1) from n=1 to infinity]
= x [d(sum x^n from n=0 to infinity) / dx]
= x [d(1/(1-x))/dx]
= x/[(1-x)^2]

👍 0 👎 0

收錄日期: 2021-04-30 01:09:36
原文連結 [永久失效]:
http://hk.knowledge.yahoo.com/question/question?qid=7009021000743

檢視 Wayback Machine 備份