由原式得a1+2a2+3a3```+nan=nSn-Sn+2n
a1+2a2+3a3```+nan+(n+1)a(n+1)=nS(n+1)+2n+2
∴nSn-Sn+2n+(n+1)a(n+1)=nS(n+1)+2n+2 (注意a1+2a2+3a3```+nan=nSn-Sn+2n
a1+2a2+3a3```+nan+(n+1)a(n+1)=nS(n+1)+2n+2
nSn-Sn+2n+(n+1)a(n+1)=nS(n+1)+2n+2
就是通过2个式子的转化)
继续.
因为 S(n+1)-Sn=a(n+1)
-Sn+(n+1)a(n+1)=n(Sn+1-Sn)+2=na(n+1)+2
所以a(n+1)=Sn+2
∴an=Sn+2
∵ Sn-S(n-1)=an 上面2个式相减得到
a(n+1)=2an
∴an是公比为2的等比数列
由题可得a1=2 ∴Sn=2^(n+1)-2
即Sn+2=2^(n+1)
所以Sn+2 等比数列