对任意的a,b∈R都有f(ab)=af(b)+bf(a),且f(0)=0,f(1)=0,f(2)=2,则f(2^-n)=_______
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![对任意的a,b∈R都有f(ab)=af(b)+bf(a),且f(0)=0,f(1)=0,f(2)=2,则f(2^-n)=_______](/uploads/image/z/5576507-35-7.jpg?t=%E5%AF%B9%E4%BB%BB%E6%84%8F%E7%9A%84a%2Cb%E2%88%88R%E9%83%BD%E6%9C%89f%28ab%29%3Daf%28b%29%2Bbf%28a%29%2C%E4%B8%94f%280%29%3D0%2Cf%281%29%3D0%2Cf%282%29%3D2%2C%E5%88%99f%282%5E-n%29%3D_______)
对任意的a,b∈R都有f(ab)=af(b)+bf(a),且f(0)=0,f(1)=0,f(2)=2,则f(2^-n)=_______
对任意的a,b∈R都有f(ab)=af(b)+bf(a),且f(0)=0,f(1)=0,f(2)=2,则f(2^-n)=_______
对任意的a,b∈R都有f(ab)=af(b)+bf(a),且f(0)=0,f(1)=0,f(2)=2,则f(2^-n)=_______
f(2^-n)=f(2^-n+1 х2)
=2^-n+1 f(2)+2f(2^-n+1)
=2^-n+2f(2^-n+2 х2)
=2^-n+2^-n+4f(2^-n+3 х2)
剩下就自己找规律罗,很简单了