Question

设函数f（x）=1/(2^X+根号2)，求f（-4）+f（-3）+...+f（5）我要知道怎么做的

像这类题有没有通性？

Answer 1

解：
f(x)=1/(2^x+√2)
f(-4)+f(-3)+……+f(5)
=1/[2^(-4)+√2]+1/[2^(-3)+√2]+1/[2^(-2)+√2]+1/[2^(-1)+√2]+1/(2^0+√2)+1/(2^1+√2)+1/(2^2+√2)+
+1/(2^3+√2)+1/(2^4+√2)+1/(2^5+√2)
=16/(1+16√2)+8/(1+8√2)+4/(1+4√2)+2/(1+2√2)+1/(1+√2)+1/(2+√2)+1/(4+√2)++1/(8+√2)+
+1/(16+√2)+1/(32+√2)
=(16-256√2)/(-511)+(8-64√2)/(-127)+(4-16√2)/(-31)+(2-4√2)/(-7)+(1-√2)/(-1)+(2-√2)/2+
+(4-√2)/14+(8-√2)/62+(16-√2)/254+(32-√2)/1022
=(256/511)√2-16/511+(64/127)√2-8/127+(16/31)√2-4/31+(4/7)√2-2/7+√2-1+1-(√2)/2+2/7-(√2)/14+
+4/31-(√2)/62+8/127-(√2)/254+16/511-(√2)/1022
=(256/511+64/127+16/31+4/7+1/2-1/14-1/62-1/254-1/1022)(√2)-16/511-8/127-4/31-
-2/7+2/7+4/31+8/127+16/511
=[1/2+(8/14-1/14)+(32/62-1/62)+(128/254-1/254)+(512/1022-1/1022)](√2)
=(1/2+1/2+1/2+1/2+1/2)(√2)
=(5/2)√2

追问

http://zhidao.baidu.com/question/232928487.html这个。。看不懂

追答

把数代进去，一步一步傻算。
傻算的过程中，注意化简。