(1).bn=1/2*(n^2+1)
(2).cn=1/[n*(n+1)]*[2^(-n-1)]*(n+1/2+bn)=1/2^(n+2)+1/2^(n+2)*[1/n+1/n*(n+1)]
=1/2^(n+2)+[1/2^(n+1)*1/n-1/2^(n+2)*1/(n+1)]=dn+cn
Sdn是等比数列好算
把cn的前n项相加可减到很多项,剩余1/2^(n+2)*1/(n+1),1/4这两项
你再算一遍吧
(1).bn=1/2*(n^2+1)
(2).cn=1/[n*(n+1)]*[2^(-n-1)]*(n+1/2+bn)=1/2^(n+2)+1/2^(n+2)*[1/n+1/n*(n+1)]
=1/2^(n+2)+[1/2^(n+1)*1/n-1/2^(n+2)*1/(n+1)]=dn+cn
Sdn是等比数列好算
把cn的前n项相加可减到很多项,剩余1/2^(n+2)*1/(n+1),1/4这两项
你再算一遍吧