這是等比數列啊

S(n) = Σ a^n = a^1 + a^2 + a^3 + ... + a^n

公比r = a^2/a^1 = a^3/a^2 = a

所以通項an = ar^(n - 1)

S(n) = a + ar + ar^2 + ar^3 + ... + a^(n - 2) + a^(n - 1)

r * S(n) = ar + ar^2 + ar^3 + ... + ar^(n - 2) + a^(n - 1) + ar^n

_____________________________________________________

相減得：S(n) - r * S(n) = a + 0 + 0 + 0 + ... - ar^n

(1 - r) * S(n) = a - ar^n = a(1 - r^n)

S(n) = a(1 - r^n)/(1 - r)，r ≠ 1

所以a^n的和式就是a(1 - r^n)/(1 - r)，r

a(r^n - 1)/(r - 1)，r > 1

2^n的公式照代入可以了