Question

设等比数列{an}的前n项和为Sn,S4=10,S12=130,求S16=

Answer 1

∵{an}为等比数列

∴Sn=a1*(1-q^n)/(1-q)
∵S4=a1*(1-q^4)/(1-q)=10
S12=a1*(1-q^12)/(1-q)=130
∴(1-q^4)/(1-q^12)=1/13
设q^4=t
∴(1-t)/(1-t³)=(1-t)/((1-t)(1+t+t²))=1/(1+t+t²)=1/13
∴1+t+t²=13
(t+4)(t-3)=0
∵t=q^4≥0
∴t=3或者t=-4（舍去）
∴a1*(1-q^4)/(1-q)=a1/(1-q)*(1-3)=10
∴a1/(1-q)=-5
∴S16=a1*(1-q^16)/(1-q)=a1/(1-q)*(1-q^16)=(-5)*(1-3^4)=5*80=400为所求

Answer 2

设第五到第九项的和为x，则有x/10=(130-10-x)/x,解得x=30，故公比为3，则s16=400