Question

关于等比数列的3道数学题，求解题过程！

Answer 1

分类: 教育/科学 >> 学习帮助
问题描述:

⒈在首项为3的等比数列{an}中，若第n项是48，第2n - 3项是192，求n？

⒉在公比为整数的等比数列{an}中，如果a1 + a4 = 18，a2 + a3 = 12，那么该数列的前8项之和是多少？

⒊在等比数列{an}中，Sn = 93，an = 48，公比q = 2，则项数n等于多少？

解析:

1:3*x^(n-1)=48;3*x^(2n-3-1)=192

=>x^(n-3)=4;x^(n-1)=16

=>x^2=4=>x=2 or x=-2

=>3*2^(n-1)=48 or 3*(-2)^(n-1)=48

=>n=5

2:a1+a4=a1+a1*x^3=a1(1+x^3)=18;a2+a3=a1(x+x^2)=12

=>(x+x^3)/(x+x^2)=18/12=3/2

=>(1+x)x/(1+x)(1-x+x^2)=3/2

=>1+x=0 or x/(1-x+x^2)=3/2

=>x=-1 or x=2 or x=1/2

=>x is integer=>x=-1 or x=2

=>if x=-1 then a1(1+(-1)^3)=18=>It is wrong.

=>if x=2 then a1(1+2^3)=18=>a1=2

=>a1+a2+...+a8=2*(1-2^8)/(1-2)=510

3:Sn=a1*(1-x^n)/(1-x)=>93=a1*(2^n-1);an=48=>a1*2^(n-1)=48

=>a1*x^n=96=>a1=96/2^n

=>93=96*(2^n-1)/2^n=96-96/2^n

=>96/2^n=3=>2^n=32=>n=5