We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive pseudonymised information about your use of our website. cookie policy and privacy policy.
 
+0  
 
0
82
2
avatar

Hello, I have been stuck on this problem: 

\(Q(n)=n /\sqrt{n} \)

Where n=(1,2,3,4.....) 

DetermineProof all positive integers n  for which:      Q(n) > Q(n+1) 

 

My thought:

By plugging in few numbers,1,2,3 etc

I noticed that Q(n)=sqrt(1),sqrt(2) etc.. 

Also that Q(1)= 1 

Where in Q(n+1) if i plugged n=1

Q(2)=1

What i am trying to say but not sure how to write the proof or the idea mathematically: that Q(n+1) sequence starts at Q(2) so the sum of its sequences must be lower than Q(n) because Q(n) starts at Q(1) 

For instance:

Q(1)=1  in q(n)

Q(2)=1 in q(n+1)

 Sep 19, 2019
 #1
avatar
0

Or did i do a mistake that Q(2)=1..

Q(2)=sqrt(2) 

But the last statement is also correct in the case too

That Q(n+1) sequence starts at Q(2) 

while Q(n) starts at Q(1) 

so the sum of Q(n) terms must be > Q(n+1) terms..

help please asap

 Sep 19, 2019
 #2
avatar+104876 
+2

n / √n  >   [ n + 1 ]  / √[n + 1]

 

n / √n   - [ n + 1 ] / √[n + 1]   > 0

 

n √[ n + 1] - [n + 1]√n 

__________________  >  0

          √n * √[n + 1]

 

 

n √[n + 1]  - [n + 1]√n  >  0       

 

n √[n + 1]  > [ n + 1]√n        square both sides

 

n^2 [ n + 1]  > [n + 1]^2 * n

 

n^2 [ n + 1] - [n + 1]^2*n > 0

 

n [n + 1]  [ n - (n + 1 ) ]  > 0

 

n (n + 1] [ - 1] >  0

 

n ( n + 1) < 0

 

This is only true on   -1 < n < 0

 

So....no poistive  integers make this true

 

Also.....see the graph here : https://www.desmos.com/calculator/tmsohselsc

 

Q(n)  is never greater than Q(n + 1)

 

 

cool cool cool

 Sep 19, 2019

20 Online Users