设数列{an}中,a1=1,且n大于1时,2Sn^2=2anSn—an求an看了答案之后卡在了1/(Sn+1)-1/Sn=2,1/S(1)=1/A(1)=1/2为首项,为什么代n=1进去呢?,2Sn^2=2anSn—an规定了n大于1啊?
问题描述：
设数列{an}中,a1=1,且n大于1时,2Sn^2=2anSn—an求an
看了答案之后卡在了1/(Sn+1)-1/Sn=2,1/S(1)=1/A(1)=1/2为首项,为什么代n=1进去呢?,2Sn^2=2anSn—an规定了n大于1啊?
答
楼主问的对.
那个答案有问题...
既然n>1,则只能说:
{1/s(n+1)}为首项=1/s(2),公差为2的等差数列.
而
a(1)=s(1)=1,
s(2)=1+a(2),a(2)=s(2)-1.
2[s(2)]^2=2a(2)s(2)-a(2)=[s(2)-1][2s(2)-1]=2[s(2)]^2-3s(2)+1,
0=1-3s(2),
s(2)=1/3.
答
an=sn-s(n-1),2sn^2=2(sn-sn-1)sn-sn+s(n-1)=2sn^2-2s(n-1)sn-sn+s(n-1) 2sns(n-1)=s(n-1)-sn
2=1/sn-1/s(n-1) n=2,2=1/s2-1/s1,n=3,2=1/s3-1/s2,n=4,2=1/s4-1/s3,.2=1/sn-1/s(n-1),相加即得1/sn-1/s1=2(n-1),又a1=s1=1,所以1/sn=2(n-1)+1/2=2n-3/2,sn=1/(2n-3/2)=2/(4n-3).(n>1)时an=sn-s(n-1)=2/(4n-3)-2/(4n-7) ．当n=1时,a1=2