Question

高一数学 等比数列

等比数列{an}中，公比q=1/2,且a3+a6+a9+···+a99=60，则a1+a2+a3+···+a99的值为
A.300
B.420
C.90
D.100

Answer 1

设a1=aq，则an=a*q^n,则a3+a6+a9+....+a99,即公比为q^3的99/3=33项的数列之和，
其和=(aq^3-aq^102)/(1-q^3)=aq^3(1-q^99)/(1-q^3)=60
而{an}中S99=(aq-aq^100)/(1-q)=aq(1-q^99)/(1-q)
两式相除得：60/S99=q^2(1-q)/(1-q^3)=1/8/(7/8)=1/7
所以S99=60*7=420
选B

Answer 2

a1+a4......+a97
=a3/q^2+a6/q^2......+a99/q^2
=4(a3 +4a6......a99)
a2+a5......+a98
=2(a3 +4a6......a99)
a1+a2+a3+···+a99=7 ( a3+a6+a9+···+a99 )=420

Answer 3

a(n)=a(1/2)^(n-1),
a(3n)=a(1/2)^(3n-1) = (a/4)(1/2)^(3n-3) = (a/4)(1/8)^(n-1),
a(3)+a(6)+...+a(3n) = (a/4)[1-(1/8)^n]/[1-1/8] = (2a/7)[1-(1/8)^n]

60 = a(3*1)+a(3*2)+...+a(3*33) = (2a/7)[1 - (1/8)^(33)],

a(1)+a(2)+...+a(99) = a[1-(1/2)^(99)]/[1-1/2] = 2a[1-(1/2)^(99)] = 2a[1-(1/8)^(33)]
= (2a/7)[1-(1/8)^(33)]*(7)
= 60*(7)
= 420

选B。

Answer 4

a1+a2+a3+.....+a99=
a1+a4+a7+....+a97+
a2+a5+a8+....+a98+
a3+a6+a9+....+a99=
4(a1/4+a4/4+a7/4+....+a97/4)+
2(a2/2+a5/2+a8/2+....+a98/2)+
a3+a6+a9+....+a99=
=4*60+2*60+60
=420
选 B