Question

求解 详细过程谢谢

Answer 1

n[S(n+1)-Sn]=Sn+n(n+1)
na(n+1)=Sn+n(n+1)
Sn=n[a(n+1)-n-1]
S(n-1)=(n-1)[an-n+1-1]=(n-1)(an-n)
Sn=S(n-1)+an=an+(n-1)(an-n)=n[a(n+1)-n-1]
nan-n²+n=na（n+1）-n²-n
n[a(n+1)-an]=2n,
a(n+1)-an=2
等差数列，d=2，a1=1；
（1）an=1+（n-1）×2=2n-1；
（2）bn=[n(n+2)(2n-1)+1]/[(n+1)(n-1)]=[n(2n²+3n-2)+1]/[(n+1)(n-1)]
=[2n³+3n²-2n+1]/[n²-1]
=2n+3+4/[(n+1)(n-1)]
=2n+3+2[1/(n-1)-1/(n+1)]
Tn=2[2+3+...+n]+3(n-1)+2[(1/1-1/3)+(1/2-1/4)+(1/3-1/5)+(1/4-1/6)+...+1/(n-1)-1/(n+1)]
=2(n+2)(n-1)/2+3(n-1)+2[1+1/2-1/n-1/(n+1)]
=(n+5)(n-1)+3-2/n-2/(n+1)

追问

写的看不懂

有手写的吗

Answer 2

向左转|向右转供参考。