设正项数列{an}前项和Sn满足Sn=1/4(an+1)^2 求an

显示全部楼层 · 2010-11-17 11:24:44

4Sn=(an+1)^2，n=1时，4S1=(a1+1)^2，4a1=(a1+1)^2，a1=1.4Sn=(an+1)^2, 4S(n-1)=[a(n-1)+1]^2,又an=Sn-S(n-1)上面两式相减得：4an=(an+1)^2-[a(n-1)+1]^2(an+1)^2-4an=[a(n-1)+1]^2(an-1)^2=[a(n-1)+1]^2，(an-1)^2-[a(n-1)+1]^2=0，（an+a(n-1)）(an-a(n-1)-2)=0,∵an>0∴an=a(n-1)+2所以an是等差数列,首项为1公差为2.an=1+2(n-1)=2n-1.

千问 · 2010-11-17 11:24:44

S1=a1=1/4(1+a1)^2==> a1=1S2=a1+a2=1/4(1+a2)^2 ==> a2=2(an>0)Sn=an+Sn-1=an+1/4[a(n-1)+1]^2 =1/4(an+1)^2==>an=1/2a(n-1)+3/2==>an=3-2*(1/2)^(n-1)

千问 · 2010-11-17 11:24:44

是An+1，还是第n+1项