设函数f(x)=1/(2^x+根号2) 则f(-2)+f(-1)+f(0)+f(1)+f(2)+f(3)=
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设函数f(x)=1/(2^x+根号2) 则f(-2)+f(-1)+f(0)+f(1)+f(2)+f(3)=
![设函数f(x)=1/(2^x+根号2) 则f(-2)+f(-1)+f(0)+f(1)+f(2)+f(3)=](/uploads/image/z/2046333-21-3.jpg?t=%E8%AE%BE%E5%87%BD%E6%95%B0f%EF%BC%88x%EF%BC%89%3D1%2F%282%5Ex%2B%E6%A0%B9%E5%8F%B72%EF%BC%89+%E5%88%99f%EF%BC%88-2%EF%BC%89%2Bf%EF%BC%88-1%EF%BC%89%2Bf%EF%BC%880%EF%BC%89%2Bf%EF%BC%881%EF%BC%89%2Bf%EF%BC%882%EF%BC%89%2Bf%EF%BC%883%EF%BC%89%3D)
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![](http://img.wesiedu.com/upload/0/45/04531c10958dd510760f08a45902293f.jpg)
所以
f(-2)+f(-1)+f(0)+f(1)+f(2)+f(3)
=[f(-2)+f(3)]+[f(-1)+f(2)]+[f(0)+f(1)]
=1/√2+1/√2+1/√2
=3/√2
=3√2/2
![](http://img.wesiedu.com/upload/0/45/04531c10958dd510760f08a45902293f.jpg)
所以
f(-2)+f(-1)+f(0)+f(1)+f(2)+f(3)
=[f(-2)+f(3)]+[f(-1)+f(2)]+[f(0)+f(1)]
=1/√2+1/√2+1/√2
=3/√2
=3√2/2
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