+4622 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 \(\boxed{2009}-\boxed{2008}\)?
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
+129899 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
+2446 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!
