Question

等比数列an的首项为a,公比为q ,Sn 为其前n项和,求S1+S2+...+Sn

Answer 1

Sn=a(1-q^n)/(1-q),

所以S1+S2+...+Sn=a(1-q+1-q^2+……+1-q^n)/(1-q)

=a[n-(q+q^2+……+q^n)]/(1-q)

由于q+q^2+……+q^n=q(1-q^n)/(1-q),代入上式,

得S1+S2+...+Sn=a[n-q(1-q^n)/(1-q)]/(1-q)

=a[n-nq-q-q^(n+1)]/(1-q)^2