Question

已知數(shù)列{a_n}、{b_n}與函數(shù)f(x)、g(x),x∈R滿足條件:b₁=b,a_n=f(b_n)=g(b_n+1)(n∈N^*).

若函數(shù)y=f(x)為R上的增函數(shù),g(x)=f^-1(x),b=1,f(1)＜1,證明對任意的n∈N^*,a_n+1＜a_n.

Answer 1

證明：因為g(x)=f^-1(x),所以a_n=g(b_n+1)=f^-1(b_n+1),即b_n+1=f(a_n).

下面用數(shù)學(xué)歸納法證明a_n+1＜a_n(n∈N^*).

(1)當(dāng)n=1時,由f(x)為增函數(shù),且f(1)＜1,得a₁=f(b₁)=f(1)＜1,

b₂=f(a₁)＜f(1)＜1,

a₂=f(b₂)＜f(1)=a₁,

即a₂＜a₁,結(jié)論成立.

(2)假設(shè)n=k時結(jié)論成立,即a_k+1＜a_k.

由f(x)為增函數(shù),得f(a_k+1)＜f(a_k),即b_k+2＜b_k+1,

進(jìn)而得f(b_k+2)＜f(b_k+1),即a_k+2＜a_k+1.

這就是說當(dāng)n=k+1時,結(jié)論也成立.

根據(jù)(1)和(2),可知對任意的n∈N^*,a_n+1＜a_n.