M&S problem

[匿名]

2011-01-09 6:48 am

Please help me to solve this :

Express 1 + 3/1! + 5/2! + 7/3! + ... in ∑ notation.

How can the ans be
圖片參考：http://imgcld.yimg.com/8/n/HA00797630/o/701101080163413873437890.jpg
?

Thx!!

更新1:

Please dun post irrelevant ans.

更新2:

But how do think of the [2(0) + 1], [2(1) + 1], [2(2) + 1], etc? Is it by the equation a(n - 1)d ??

更新3:

意思是說如果n由0開始的話代n + 1 入 n 就可以了?

回答 (1)

用戶2053

2011-01-17 2:36 am

✔ 最佳答案

1 + 3/1! + 5/2! + 7/3! + ...= 1/0! + 3/1! + 5/2! + 7/3! + ...= [2(0) + 1]/0! + [2(1) + 1]/1! + [2(2) + 1]/2! + ...=
圖片參考：http://imgcld.yimg.com/8/n/HA04628698/o/701101080163413873437900.jpg

2011-01-16 21:59:15 補充：
Yes , it's like to find the genernal term :

1 , 3 , 5 , 7 is a.s. ,

T(1) = 1 = 2*1 - 1
T(2) = 3 = 2*2 - 1
T(3) = 5 = 2*3 - 1
........
T(n) = 2n - 1

So

The answer also can be written as

∞
Σ (2n - 1)/(n - 1)!
n=1

2011-01-16 21:59:22 補充：
But here n is start from 0 , it is easy to change the above form by
sub N+1 to n :

(2(N+1) - 1) / (N+1 - 1)!

= (2N + 1) / N!

So

∞
Σ (2n + 1)/n!
n=0

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