1/k+4/k^2+9/k^3+16/k^4+... (k>1).
參考:1/k+2/k^2+3/k^3+4/k^4+... =k/(k-1)^2, (k>1)
1/k+4/k^2+9/k^3+16/k^4+... (k>1).
參考:1/k+2/k^2+3/k^3+4/k^4+... =k/(k-1)^2, (k>1)
•
對已知導數兩邊求導即可
-魁北克人-
♂
(631 bytes)
()
11/11/2018 postreply
18:05:20
•
我想了半天才明白您的意思,謝謝!
-yma16-
♂
(0 bytes)
()
11/12/2018 postreply
21:48:57
•
按照您的思路,要是把分子換成立方,也可以做。
-yma16-
♂
(0 bytes)
()
11/12/2018 postreply
21:53:07
•
直接的解法
-yma16-
♂
(2760 bytes)
()
11/13/2018 postreply
19:31:19
WENXUECITY.COM does not represent or guarantee the truthfulness, accuracy, or reliability of any of communications posted by other users.
Copyright ©1998-2025 wenxuecity.com All rights reserved. Privacy Statement & Terms of Use & User Privacy Protection Policy