- 跪求帮助~~~~~~已知数列an的各项均为正数,前n项和为Sn,
- 已知数列an的各项均为正数,前n项和为Sn,且满足2Sn=an^2+n-4
(1)求证an为等差数列
(2)设bn=2^an,求bn前n项的和
- (1)
2Sn=an^2+n-4(1)
2S=a^2+(n-1)-4=a^2+n-5(2)
(1)-(2) 2Sn-2S=an^2+n-4-(a^2+n-5)
2an=an^2-a^2+1
a^2=(an-1)^2
各项为正数,所以a=an-1,
an-a=1, {an}为等差数列
(2)
等差数列{an}中,2a1=a1^2+1-4, a1^2-2a1-3=0
a1为正数,a1=3, 又公差d=1
an=3+(n-1)=2+n
bn=2^(an)=2^(2+n)=4*2^n
{bn}是公比为2的等比数列,b1=8,
前n项和=8(2^n-1)/(2-1)=8(2^n-1)