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Loser66
 one year ago
Conjecture the formula for the nth term of {a_n} if the first ten terms of it is
1,0,0,1,0,0,0,0,1,0
Please, help
Loser66
 one year ago
Conjecture the formula for the nth term of {a_n} if the first ten terms of it is 1,0,0,1,0,0,0,0,1,0 Please, help

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Loser66
 one year ago
Best ResponseYou've already chosen the best response.1I want to use the floor function to put it in neat. Obviously, dw:1440875326682:dw

freckles
 one year ago
Best ResponseYou've already chosen the best response.0what do you mean how if you say it is obvious?

freckles
 one year ago
Best ResponseYou've already chosen the best response.0is the piecewise part obvious but not how to write in the floor format? is that what you?

freckles
 one year ago
Best ResponseYou've already chosen the best response.0is that what you mean*?

ikram002p
 one year ago
Best ResponseYou've already chosen the best response.2oh that remind me of Legendre function

Loser66
 one year ago
Best ResponseYou've already chosen the best response.1When I say "obvious" , it means we can see it obviously.

freckles
 one year ago
Best ResponseYou've already chosen the best response.0ok then what do you need help on then

Loser66
 one year ago
Best ResponseYou've already chosen the best response.1I want the floor format.

Loser66
 one year ago
Best ResponseYou've already chosen the best response.1\(\lfloor{\sqrt n}\rfloor\)

Loser66
 one year ago
Best ResponseYou've already chosen the best response.1How about this? \(1<\sqrt 2 < 2\), hence \(\lfloor{\sqrt 2}\rfloor =1\) and \(\lfloor{\dfrac{\lfloor{\sqrt 2}\rfloor}{\sqrt2}}\rfloor=0\)

Loser66
 one year ago
Best ResponseYou've already chosen the best response.1It works well for n =3,4, 5, 6... but how can I just list them out like this? I want a general logic to lead me there.

ikram002p
 one year ago
Best ResponseYou've already chosen the best response.2so sqrt(n^2) is integer, then [sqrt(n^2) ] would be same integer n else u would have an additional decimals there are two lovely things about floor function 1[sqrt(n^2) ]<= sqrt(n^2) 2[r]=0 if 0<r<1 and you already know [sqrt(n^2) ]<= sqrt(n^2) \(\dfrac {[\sqrt(n^2) ]}{\sqrt(n^2)}<=1\) we have two cases case 1 \(\dfrac {[\sqrt(n^2) ]}{\sqrt(n^2)}=1\) which mean n is an integer and n^2 is a square of integer also \([\dfrac {[\sqrt(n^2) ]}{\sqrt(n^2)}] =1\) case 2 \(\dfrac {[\sqrt(n^2) ]}{\sqrt(n^2)}<1\) from property 2 \([\dfrac {[\sqrt(n^2) ]}{\sqrt(n^2)}] =0 \) which u already mean that n is not a square of integer

ikram002p
 one year ago
Best ResponseYou've already chosen the best response.2and you can use n =i^2 instead n^2 as the general term lol

Loser66
 one year ago
Best ResponseYou've already chosen the best response.1Hey, friend , I don't get why \(\sqrt n^2 \). Is it not that it is always = n and n >0 since it is the order of the term.

Loser66
 one year ago
Best ResponseYou've already chosen the best response.1If we consider \(\lfloor{\sqrt n^2}\rfloor = \lfloor{n}\rfloor\) and it is = n itself.

Loser66
 one year ago
Best ResponseYou've already chosen the best response.1since \(n\in \mathbb N\)

ikram002p
 one year ago
Best ResponseYou've already chosen the best response.2yes u are correct , its just a variable :D sorry to confuse u ;)

ikram002p
 one year ago
Best ResponseYou've already chosen the best response.2i just used it to explain but since u were asked to make general form of n then use n without n^2

Loser66
 one year ago
Best ResponseYou've already chosen the best response.1oh, I got you. Thank you so much.

ikram002p
 one year ago
Best ResponseYou've already chosen the best response.2you are the most wlc <3
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