已知函数f(x+2/1)为奇函数,设g(x)＝f(x)＋1,则g(1/2012)＋g(2/2012)+.+g(2011/2012)＝答案是2011.

Question

已知函数f(x+2/1)为奇函数,设g(x)＝f(x)＋1,则g(1/2012)＋g(2/2012)+.+g(2011/2012)＝____
答案是2011.

Answer 1

因为,f(x+1/2)为奇函数
所以,f(-x+1/2)=-f(x+1/2)
f(1/2012)＝f(-1005/2012＋1/2)＝-f(1005/2012＋1/2)＝-f(2011/2012)
f(2/2012)＝f(-1004/2012＋1/2)＝-f(1004/2012＋1/2)＝-f(2010/2012)
……
f(1005/2012)＝f(-1/2012＋1/2)＝-f(1/2012＋1/2)＝-f(1007/2012)
而,f(1006/2012)＝f(0＋1/2)＝-f(0＋1/2)＝-f(1006/2012)
即,f(1006/2012)=f(1/2)＝0
所以,f(1/2012)＋f(2/2012)+.+f(2011/2012)＝0
g(x)＝f(x)＋1
g(1/2012)＋g(2/2012)+.+g(2011/2012)
＝f(1/2012)＋f(2/2012)+.+f(2011/2012)＋2011
＝0＋2011
＝2011