Question

an前n项和为sn,已知a1=1,a2=2,且an+2=3sn-sn+1+3, (1)证明an+2=3an(n+2,n为下标）， (2

an前n项和为sn,已知a1=1,a2=2,且an+2=3sn-sn+1+3, (1)证明an+2=3an(n+2,n为下标）， (2）求Sn

Answer 1

解：(1)a2²=S2+S1=a1+a2+a1=2a1+a2=2×1+a2=a2+2a2²-a2-2=0(a2+1)(a2-2)=0a2=-1(舍去)或a2=2a(n+1)²=S(n+1)+Sna(n+2)²=S(n+2)+S(n+1)a(n+2)²-a(n+1)²=S(n+2)-Sn=a(n+2)+a(n+1)[a(n+2)+a(n+1)][a(n+2)-a(n+1)]-[a(n+2)+a(n+1)]=0[a(n+2)+a(n+1)][a(n+2)-a(n+1)-1]=0数列是正项数列，a(n+2)+a(n+1)恒>0，因此只有a(n+2)-a(n+1)-1=0a(n+2)-a(n+1)=1，为定值，又a2-a1=2-1=1，数列{an}是以1为首项，1为公差的等差数列。an=1+1×(n-1)=nn=1时，a1=1，同样满足表达式数列{an}的通项公式为an=n(2)bn=a(2n-1)·2^(an)=(2n-1)·2ⁿTn=1·2+3·2²+5·2³++(2n-1)·2ⁿ2Tn=1·2²+3·2³++(2n-3)·2ⁿ+(2n-1)·2ⁿ⁺¹Tn-2Tn=-Tn=2+2·2²+2·2³++2·2ⁿ-(2n-1)·2ⁿ⁺¹=2·(2+2²++2ⁿ)-(2n-1)·2ⁿ⁺¹-2=2·2·(2ⁿ-1)/(2-1)-(2n-1)·2ⁿ⁺¹-2=(3-2n)·2ⁿ⁺¹+6Tn=(2n-3)·2ⁿ⁺¹+6

Answer 2

简单分析一下，答案如图所示