f(x)=x²tan2a+xsin(2a+π/4)=x^2+x
a(n+1)=f(an)=(an)^2+an=an(an+1);
兩邊取導(dǎo)
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,數(shù)列是遞增的,
由a1=1/2,所以1/2