+0

# help!

0
361
4
+4287

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
+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
+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

Feb 6, 2018
#3
+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
+101360
0

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

Feb 6, 2018