x+x^1+x^2+x^3+.....+x^n = ??

x + x^1 + x^2 + x^3 + ..... + x^n
= x + x + x^2 + x^3 + ..... + x^n
= x + (x + x^2 + x^3 + ..... + x^n)

x + x^2 + x^3 + ..... + x^n 為一等比數列的前 n 項和。
首項 a = x，公比 r = x，項數 = n
根據公式： 等比數列的前 n 項和 = a(r^n - 1)/(r - 1)

x + x^1 + x^2 + x^3 + ..... + x^n
= x + (x + x^2 + x^3 + ..... + x^n)
= x + [x(x^n - 1)/(x - 1)]
= [x(x - 1)/(x - 1)] + [x(x^n - 1)/(x - 1)]
= [x^2 - x + x^(n+1) - x] / (x - 1)
= [x^(n+1) + x^2 - 2x] / (x - 1)

x+x^1+x^2+x^3+.....+x^n = ? --------(1)
1+1+x^1+x^2+.....+x^(n-1) = ?/x ---------(2)
(1) - (2) , x^n + x - 2 = ?(1-1/x)
x^(n+1)+x^2 - 2x = ?(x-1)
=> ? = [x^(n+1)+x^2 - 2x]/(x-1)

2012-08-25 22:08:35 補充：
assuming x =/=0, otherwise, reverse (1) and (2) for the left hand side by assuming 1+1+x^2 + x^2 + ... + x^(n-1) be "?"

2012-08-25 22:09:46 補充：
Furthermore, for 土扁,
in your case, u need to assume x =/= 1,
otherwise,
x + x^1 + x^2 + x^3 + ..... + x^n
= 1+1^1+1^2+1^3+...+1^n
= n+1