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For positive integer values of \(N\), let \(\boxed N\)  be defined as \(\begin{align*} &\boxed N=2+4+6+\ldots+N\text{, if }N\text{ is even and}\\ &\boxed N=1+3+5+\ldots+N\text{, if }N\text{ is odd.} \end{align*}\) What is the value of \(\boxed{2009}-\boxed{2008}\)?

 Feb 6, 2018
 #1
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+1

N=[L - F] / D + 1

N=1,004 for EVEN terms

N =1,005 for ODD terms

Sum =[F + L] / 2 x N

Sum =1,009,020 sum of EVEN terms

Sum =1,010,025 sum of ODD terms

1,010,025 - 1,009,020 = 1,005

 Feb 6, 2018
 #2
avatar+101360 
+1

Note that the sum   of

 

2 + 4 + 6 +....+ N         can  be written as  (N + 2) * (N /4) 

 

And the sum of  

 

1 + 3  + 5 +  ....+ N     can be written as  [ (N + 1)/2]^2  =  (N + 1)^2 / 4  =

 

So  we have 

 

[ 2009 + 1 ] ^2 / 4      -  [2008 + 2] [2008] / 4 =

 

 

[2010]^2 / 4   -   [ 2010] *[ 2008] / 4  =

 

[2010] / 4  * [ 2010  - 2008]  =

 

[ 2010 ]/ 4  *  2  =

 

1005

 

 

 

cool cool cool

 Feb 6, 2018
 #3
avatar+2339 
+1

Cphill and Guest have providing their individual ways of thinking about it. I think I will, too!

 

\(\boxed{2009}=\hspace{5mm}1+3+5+...+2003+2005+2007+2009\\ \boxed{2008}=\underline{-(2+4+6+...+2004+2006+2008)}\\ \hspace{19mm}\underbrace{-1-1-1-...-1-1-1}\hspace{21mm}+2009\\ \hspace{33mm}\text{1004 times}\)

 

Therefore, we have concluded that the complicated expression has been reduced to something relatively simple.

 

\(-1*1004+2009\\ -1004+2009\\ 1005\)

 

This is exactly what everyone else got!

 Feb 6, 2018
 #4
avatar+101360 
0

I actually like your way better, X2....!!!!

 

 

cool cool cool

 Feb 6, 2018

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