an=snsn1是什么公式

sn=n^2+1,判断是否为等差数列,tn=3/s1s2+5/s2s3+...2n+1/snsn1
an=Sn-S(n-1)=n^2+1-(n-1)^2-1=2n-1a(n-1)=2n-3an-a(n-1)=2所以an为等差数列(2n+1)/[sns(n+1)]=(2n+1)/[(n^2+1)((n+1)^2+1)]=[(n+1)^2+1-(n+1)^2]/[(n^2+1)((n+1)^2+1)]=1/(n^2+1)-1/[(n+1)^2+1]=1/sn-1/s(n+1)tn=3/(s说完了。