如何求等比数列的前n项和的前n项和?

a(n) = aq^(n-1),
s(n) = a(1)+a(2)+...+a(n),
t(n) = s(1)+s(2)+...+s(n).
q=1时,a(n) = a,s(n) = na,t(n) = n(n+1)a/2.
q不为1时,a(n) = aq^(n-1),s(n) = a[1-q^n]/(1-q) = a/(1-q) - [aq/(1-q)]q^(n-1),
t(n) = na/(1-q) - [aq/(1-q)][1-q^n]/(1-q)
= na/(1-q) - aq(1-q^n)/(1-q)^2

1年前