Please help.
Find the value of 2+(2^3)/3!+(2^5)/5!+(2^7)/7!+...+(2^2r+1)/(2r+1)!+.... in terms of e.
F4 Maths: Exponential Series
2010-04-07 1:27 am
回答 (2)
2010-04-07 2:46 am
✔ 最佳答案
By the ordinary exponential series:ex = 1 + x + x2/2! + x3/3! + ...
Let k = 2 + 23/3! + 25/5! + ... + 22r+1/(2r + 1)! + ...
Sub x = 2, then
e2 = 1 + 2 + 22/2! + 23/3! + 24/4! + ...
= (1 + 22/2! + 24/4! + ...) + k
Sub x = -2, then
e-2 = 1 - 2 + 22/2! - 23/3! + 24/4! - ...
= (1 + 22/2! + 24/4! + ...) - k
Then
e2 - e-2 = 2k
k = (e2 - e-2)/2
參考: Myself
2010-04-07 2:43 am
Hope I can help you !
Module 2 : Algebra and calculus
2+(2^3)/3!+(2^5)/5!+(2^7)/7!+...+(2^2r+1)/(2r+1)!+.... = ?
Answer :
Since , e^x = 1 + x + x²/2! + x³/3! + .......
Let x = 2 ,
e^2 = 1 + 2 + 2²/2! + 2³/3! + ....... (1)
Considerate (1) and[ 2+(2^3)/3!+(2^5)/5!+(2^7)/7!+...+(2^2r+1)/(2r+1)!+....]
2+(2^3)/3!+(2^5)/5!+(2^7)/7!+...+(2^2r+1)/(2r+1)!+....
= (1) - [ 1+ 2²/2! + (2^4)/4! + ....]
= e^2 - (1+2+16/24+4/45+2/315+....+(2^2r)/(2r)!+...)
= ...........
Module 2 : Algebra and calculus
2+(2^3)/3!+(2^5)/5!+(2^7)/7!+...+(2^2r+1)/(2r+1)!+.... = ?
Answer :
Since , e^x = 1 + x + x²/2! + x³/3! + .......
Let x = 2 ,
e^2 = 1 + 2 + 2²/2! + 2³/3! + ....... (1)
Considerate (1) and[ 2+(2^3)/3!+(2^5)/5!+(2^7)/7!+...+(2^2r+1)/(2r+1)!+....]
2+(2^3)/3!+(2^5)/5!+(2^7)/7!+...+(2^2r+1)/(2r+1)!+....
= (1) - [ 1+ 2²/2! + (2^4)/4! + ....]
= e^2 - (1+2+16/24+4/45+2/315+....+(2^2r)/(2r)!+...)
= ...........
參考: By myself
收錄日期: 2021-04-16 12:16:32
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