已知正项数列{an}的前n项和为Sn,且a1=1, a²n+1=Sn+1+Sn 求{an}的通项公式
求{an}的通项公式
设bn=a2n-1·2^an,求数列{bn}的前n项和Tn 展开
解:
(1)
a2²=S2+S1=a1+a2+a1=2a1+a2=2×1+a2=a2+2
a2²-a2-2=0
(a2+1)(a2-2)=0
a2=-1(舍去)或a2=2
a(n+1)²=S(n+1)+Sn
a(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=0
a(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)=n
n=1时,a1=1,同样姿指满足表达纯册悉式
数列{an}的通项公式为an=n
(2)
bn=a(2n-1)·2^(an)=(2n-1)·2ⁿ
Tn=1·2+3·2²+5·2³+...+(2n-1)·2ⁿ
2Tn=1·2²+3·2³+...+(2n-3)·2ⁿ+(2n-1)·2ⁿ⁺¹
Tn-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做乎⁺¹+6
Tn=(2n-3)·2ⁿ⁺¹+6