x=an f(x)=a(n+1)代入函數(shù)方程
a(n+1)=an^2 +2an
a(n+1) +1=an^2+2an +1=(an +1)^2
滿足平方遞推數(shù)列定義,因此數(shù)列{an +1}是平方遞推數(shù)列.
a1+1=10>0,若當(dāng)n=k(k∈N+)時(shí),ak>0,則a(k+1)=ak^2>0,k為任意正整數(shù),因此對(duì)于任意正整數(shù)n,數(shù)列{an +1}各項(xiàng)恒為正.
lg(a1+1)=lg(9+1)=lg10=1
an +1=[a(n-1)+1]^2
=[a(n-2)+1]^(2^2)
=[a(n-3)+1]^(2^3)
=.
=[a1+1]^[2^(n-1)]
=(9+1)^[2^(n-1)]
=10^[2^(n-1)]
lg(an +1)=lg[10^[2^(n-1)]]=2^(n-1)
lg[a(n+1)+1]/(an +1)=2^n/2^(n-1)=2,為定值,數(shù)列{lg(an +1)}是以1為首項(xiàng),2為公比的等比數(shù)列
lg(an +1)=1×2^(n-1)=2^(n-1)
lg(Tn)=lg[(a1+1)(a2+1)...(an +1)]
=lg(a1+1)+lg(a2+1)+...+lg(an +1)
=1×(2ⁿ -1)/(2-1) /這一步不難理解吧,就是等比數(shù)列求和
=2ⁿ-1