Find \(2000\) th term of below sequence \[2,3,5,6,7,8,10,\ldots\] (the sequence contains positive integers other than perfect squares)
\(\huge\color{white}{{ *}}\)
993
my bad
1993
sequence of a number that are not squares
thats very close ! but it has to be greater than 2000 for sure... as the 2000th term of positive integer sequence(1,2,3,4,...) itself is 2000. i am looking for answer and a neat way to justify the answer :)
the 2000th term is 2000? I do not think so. We started with 2 and skipped 4 and 9 that is down to 10 - 3 sequence
2000th term of below sequence is 2000 : \[1,2,3,4,5, \ldots\] so the 2000th term of the sequence in main question has to be greater than 2000
ohh hahahha
right
can I just give you the pseudo-code in Python or C++? laughing out loud
lol it can be considered a sledgehammer in the context of this problem :P
2044?
till 2000.. 44 squares are missing from the series last of which is 1936.. so 2044
thats very close too.. but no!
then 2045?
2045 is right ! :)
how do I interpret this? \(a_n = \lfloor \sqrt{n}+\frac{1}{2})\rfloor +n\)
that looks interesting xD is that the general term of sequence ?
it is, apparently
don't mind me, I am crashing from the caffeine-induced waking state
that actually works! http://www.wolframalpha.com/input/?i=floor%28sqrt%282000%29%2B1%2F2%29%2B2000
2045? 1935:1979 1956:2000 1980:2024 1981:2026 2000:2045
i thought of it like this sequence of perfect square up to some value http://www.wolframalpha.com/input/?i=n%5E2+from+n%3D1+to+n%3D50 so we can note that 2000-44=1956 so the 1956 term is 2000 now we keep moving like this 2001.. to 2024 ( the 1980 term ) now we need other 20 terms :P from 2026 .... to 2045 so i would say 2045
Suscribing to the post :P
Oops *Subscribing
oh hia der...
Yeah I like @ikram002p 's way, that's how I was gonna do it haha.
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