perplexus.info :: Just Math : Cube rooted till 2019

Cube rooted till 2019 (Posted on 2019-10-23)

Find the floor of the given expression:

1/(1)^1/3+1/(2)^1/3+1/(3)^1/3+...+1/(2019) ^1/3

No Solution Yet Submitted by Danish Ahmed Khan

No Rating

Comments: ( Back to comment list | You must be logged in to post comments.)

Puzzle Answer

Comment 4 of 4 |

The required floor value corresponding to the given expression is: 238

** I sincerely hope to posit an appropriate explanation in future.

Edited on April 25, 2022, 2:17 am

Posted by K Sengupta on 2022-04-25 02:15:44

Please log in:

Forums (0)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (0)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Chatterbox:


perplexus dot info