已知a为锐角,且tana=(根号2)-1,函数f(x)=x²tan2a+xsin(2a+π/4),数列an的首相a1=1/2,an+1=f(an)求证:1扫码下载作业帮拍照答疑一拍即得

问题描述:

已知a为锐角,且tana=(根号2)-1,函数f(x)=x²tan2a+xsin(2a+π/4),数列an的首相a1=1/2,an+1=f(an)
求证:1

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拍照答疑一拍即得

tana=√2-1tan2a=2tana/(1-tan^2a)=1,2a=π/4
f(x)=x²tan2a+xsin(2a+π/4)=x^2+x
a(n+1)=f(an)=(an)^2+an=an(an+1);
两边取导
1/a(n+1)=1/an(an+1)=1/an-1/(an+1)
所以1/(an+1)=1/an - 1/a(n+1)
求和:1/﹙1+a1﹚+1/﹙1+a2﹚+…+1/﹙1+an﹚
=1/a1-1/a2+1/a2-1/a3+...+1/an - 1/a(n+1)=1/a1-1/a(n+1)
a(n+1)=(an)^2+an,数列是递增的,
由a1=1/2,所以1/2把a1的范围带进去就得证了