+0  
 
0
55
2
avatar+454 

please explain, thank you very much!

YEEEEEET  Nov 17, 2018
 #1
avatar+316 
+4

 11) \(\frac{α}{β}> \frac{2α-β}{α+4β} \) multiply by α+4β and β and its >0 because α,β its positive numbers 

\(α(α+4β)>β(2α-β) <=> α^2+4αβ > 2αβ-β^2 <=> α^2 +4αβ - 2αβ + β^2 >0 <=> α^2 +2αβ + β^2 >0 <=> (α+β)^2>0 \)

where applicable! So we prove it!

 

12) Something is wrong here or I don't understand the question:

exp.:\(3^2-2^2=9-4=5\)  But there is no sum of two consecutive odd numbers be equal with 5 because 1+3 = 4 and general  sum of two consecutive odd numbers  is allways be equal with even number because if n is a number we have:

n+(n+2)=2n+2 = 2(n+1) for every n natural so we have 2(n+1) 

is divided by 2 so it's even number and 5(in this example) it's not even number.

 

 Hope this helps!

Dimitristhym  Nov 18, 2018
edited by Dimitristhym  Nov 18, 2018
edited by Dimitristhym  Nov 18, 2018
 #2
avatar+92785 
+2

Here's  12

 

Assuming that  we are dealing with positive integers....let the odd numbers  be

2n - 1   and 2n + 1

Adding these together we get   4n

 

Sometimes.....in these kinds of proofs, it is good to see if we can discover a pattern....

 

Let n = 2

And the sum of the odds =  8

And the difference of two squares that equals this  is 3^2 - 1^2 = 9 - 1 = 8

We can write this as  [ 2n - ]^2 - [ 2n - 3]^2

Note that the numbers in red are just  (n - 1) and (n + 1), respectively

 

Now......Let  n = 3

And the sum of the odds = 12

And the difference of two squares that equals this is 4^2 - 2^2  = 16 - 4  = 12

Which we can write as [ 2n - ]^2 - [ 2n - 4 ]^2 

Note that, once again, the numbers in red are just ( n - 1)  and  ( n + 1), respectively

 

 

So....it appears that we can write the larger  square  as   [2n - (n- 1)]^2   =  [ n + 1]^2

And it appears that we can write the other square as  [ 2n - (n + 1) ]^2  =  [ n - 1]^2

 

And their difference is

 

[ n + 1]^2  -   [ n - 1]^2   =

 

[ n^2 + 2n + 1 ]  - [ n^2 - 2n + 1]  =

 

4n

 

cool cool cool

CPhill  Nov 18, 2018
edited by CPhill  Nov 18, 2018
edited by CPhill  Nov 18, 2018
edited by CPhill  Nov 18, 2018

7 Online Users

New Privacy Policy

We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive information about your use of our website.
For more information: our cookie policy and privacy policy.