12+1122+111222+11112222+...+11...122...2(n个1,n个2）的和求12+1122+111222+11112222+...+11...122...2（n个1,n个2）的和

An=(1/9)(10^n-1)*10^n + (2/9)(10^n-1)=(1/9)[100^n+10^n-2]分组求和ok
Sn=[100^(n+1)+11*10^(n+1)-198n-210]/891

a(n) = 1111..12222..2(n个1，n个2)
= 1111..100...0(n个1，n个0) + 2222...2(n个2)
= 10^n*[1111..1(n个1)] + 2*[1111..1(n个1)]
= [2 + 10^n][1111...1(n个1)]
1111...1(n个1)
= 1 + 10 + 10^2 + ... + 10^(n-1)
= [10^n - 1]/[10-1]
= [10^n - 1]/9
所以，
a(n) = 1111..12222..2(n个1，n个2)
= [2 + 10^n][1111...1(n个1)]
= [2 + 10^n][10^n - 1]/9
= [10^n + 100^n - 2]/9
a(1) + a(2) + ... + a(n)
= [10^1 + 100^1 - 2]/9 + [10^2 + 100^2 - 2]/9 + ... + [10^n + 100^n - 2]/9
= [10 + 10^2 + ... + 10^n]/9 + [100 + 100^2 + ... + 100^n]/9 - 2n/9
= 10[10^n - 1]/81 + 100[100^n - 1]/[9*99] - 2n/9
= 10^(n+1)/81 - 10/81 + 100^(n+1)/[891] - 100/[891] - 2n/9
= 10^(n+1)/891 + 10^(n+1)/81 - 2n/9 - 70/297

因为
111..11222..22（n个1,n个2）=111..11（n个1）X100...00（n个0）+2X111..11（n个1）=（1/9)(10^n-1)*10^n + (2/9)(10^n-1)=(1/9)（100^n+10^n-2）
所以
原式=（1/9)[100^n+100^(n-1)+...+100^2+10^n+10^(n-1)+...+10^2-2n]
=[100^(n+1)+11*10^(n+1)-198n-210]/891