高二数学问题，第二题，数列求和，用的是错位相减法。。。。。我做到这一步感觉有点不对啊

s(n) = (1+1)(10/11) + (2+1)(10/11)^2 + ... + (n-1+1)(10/11)^(n-1) + (n+1)(10/11)^n,
(10/11)s(n) = (1+1)(10/11)^2 + (2+1)(10/11)^3 + ... + (n-1+1)(10/11)^n + (n+1)(10/11)^(n+1),

(1/11)s(n) = s(n) - (10/11)s(n) = (1+1)(10/11) + (10/11)^2 + ...+(10/11)^n - (n+1)(10/11)^(n+1)
= 10/11 + (10/11)[1 + (10/11) + ... + (10/11)^(n-1)]- (n+1)(10/11)^(n+1) [至此，和楼主完全一样。。楼主威武。。]
= 10/11 + (10/11)[ 1 - (10/11)^n]/(1-10/11) - (n+1)(10/11)^(n+1)
= 10/11 + 10[1 - (10/11)^n] - (n+1)(10/11)^(n+1)
= 120/11 - 11(10/11)^(n+1) - (n+1)(10/11)^(n+1)
= 120/11 - (n+12)(10/11)^(n+1),

s(n) = 120 - 11(n+12)(10/11)^(n+1).

过程没错，想多了