Question

求高手帮忙解等比数列

设等比数列{an}的前n项的和为Sn，且a1+a2+a3=351，a2+a3+a4=117求limSn n→∞

Answer 1

解：
(a2+a3+a4)/(a1+a2+a3)
=a2(1+q+q²)/[a1(1+q+q²)]
=a2/a1
=q=117/351=1/3
q=1/3
a1(1+q+q²)=351
a1(1+1/3+1/9)=351
a1=243
Sn=a1(1-q^n)/(1-q)
=243(1-1/3^n)/(2/3)
=(729/2)(1-1/3^n)
n->+∞ 1/3^n->0
1-1/3^n->1
Sn->729/2
limSn=729/2

Answer 2

a2+a3+a4=q(a1+a2+a3)
117=q*351
q=1/3
a1+a2+a3
=a1(1+q+q^2)
=a1(1+1/3+1/9)
=13a1/9
13a1/9=351
a1=243

Sn=a1(1-q^n)/(1-q)
=243[1-(1/3)^n]/(1-1/3)
=729[1-(1/3)^n]/2
=729/2-729(1/3)^n/2
=729/2-3^6*(1/3)^n/2
=729/2-(1/3)^(n-6)/2
limSn n→∞=729/2