等比數列與等比級數的問題

S = 2 + 3 + (9/2) + (27/4) + ..... + (2187/64)


S = a + ar + ar² + ar³ + ............

a = 2
r = ar / a = 3/2

2014-03-16 21:02:50 補充：
(2187/64) / (2) = r^(n-1) = (3/2)^(n-1)

(2187/128) = (3/2)^(n-1)

log(2187/128) / log(3/2) = n-1

n-1=7

n=8

2014-03-16 21:08:49 補充：
S(n) = a(1-r^n)/(1-r)

=(2)[1-(3/2)^8]/[1-(3/2)]

=(2)[1-6561/256] / [-1/2]

=(2)[-6305/256] / [-1/2]

=6305 / 64

等比公式：a1*(1-r的n次方)/(1-r)

所以

2*[1-(3/2)的8次方]/(1-3/2)

算出來=6305/64