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