已知函数y=f(x)的定义域为R,当x1,且对任意的实数x,y(x,y属于R),等式f(x)*f(y)=f(x+y)成立,若实数{an}满足a1=f(0),且f(a(n+1))=1/f(-2-an),则a2011的值为?
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![已知函数y=f(x)的定义域为R,当x1,且对任意的实数x,y(x,y属于R),等式f(x)*f(y)=f(x+y)成立,若实数{an}满足a1=f(0),且f(a(n+1))=1/f(-2-an),则a2011的值为?](/uploads/image/z/5260972-4-2.jpg?t=%E5%B7%B2%E7%9F%A5%E5%87%BD%E6%95%B0y%3Df%EF%BC%88x%EF%BC%89%E7%9A%84%E5%AE%9A%E4%B9%89%E5%9F%9F%E4%B8%BAR%2C%E5%BD%93x1%2C%E4%B8%94%E5%AF%B9%E4%BB%BB%E6%84%8F%E7%9A%84%E5%AE%9E%E6%95%B0x%2Cy%EF%BC%88x%2Cy%E5%B1%9E%E4%BA%8ER%EF%BC%89%2C%E7%AD%89%E5%BC%8Ff%EF%BC%88x%EF%BC%89%2Af%EF%BC%88y%EF%BC%89%3Df%EF%BC%88x%2By%EF%BC%89%E6%88%90%E7%AB%8B%2C%E8%8B%A5%E5%AE%9E%E6%95%B0%7Ban%7D%E6%BB%A1%E8%B6%B3a1%3Df%EF%BC%880%EF%BC%89%2C%E4%B8%94f%EF%BC%88a%EF%BC%88n%2B1%EF%BC%89%EF%BC%89%3D1%2Ff%EF%BC%88-2-an%EF%BC%89%2C%E5%88%99a2011%E7%9A%84%E5%80%BC%E4%B8%BA%3F)
已知函数y=f(x)的定义域为R,当x1,且对任意的实数x,y(x,y属于R),等式f(x)*f(y)=f(x+y)成立,若实数{an}满足a1=f(0),且f(a(n+1))=1/f(-2-an),则a2011的值为?
已知函数y=f(x)的定义域为R,当x1,且对任意的实数x,y(x,y属于R),等式f(x)*f(y)=f(x+y)成立,若实数{an}满足a1=f(0),且f(a(n+1))=1/f(-2-an),则a2011的值为?
已知函数y=f(x)的定义域为R,当x1,且对任意的实数x,y(x,y属于R),等式f(x)*f(y)=f(x+y)成立,若实数{an}满足a1=f(0),且f(a(n+1))=1/f(-2-an),则a2011的值为?
取x=-1,y=0,则f(-1)*f(0)=f(-1),由当x1,得:f(0)=1
下证函数具有单调递减性(自行证明)
f(a(n+1))=1/f(-2-an)可得:f(a(n+1))*f(-2-an)=f(a(n+1)-2-an)=f(0)
所以由单调性可得:a(n+1)-2-an=0,即a(n+1)-an=2(常数)
所以数列为等差数列,下面就简单了,a2011=1+(2011-1)*2=4021
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