Question

已知{an}是等比数列,a2=2,a5=1/4,则a1a2+a2a3+...+anan+1

已知{an}是等比数列,a2=2,a5=1/4,则a1a2+a2a3+...+anan+1=?
请求高手详解,谢高!!!!!!!!!!!!!!!!!

Answer 1

a2a3/a1a2=a3/a1=q^2
a5/a2=q^3=1/8,
q=1/2,q^2=1/4,
a1=a2/q=2/(1/2)=4,
a1a2=4*2=8,
a1a2+a2a3+...+anan+1
=8[1-(1/4)^2n]/(1-1/4)
=32[1-(1/4)^2n]/3

Answer 2

a5/a2=q^3=1/4/2=1/8
q=1/2

所以a1=a2/q=4

设B1=a1a2=a1^2q=4^2*1/2=8
B2=a2a3=a1^2q^3
...
Bn=anan+1=a1^2q^(2n-1)

此Bn乃是首项为8,公比为q^2=1/4的等比数列

a1a2+a2a3+...+anan+1=B1+B2+..+Bn=8(1-(1/4)^n)/(1-1/4)=32/3-1/3*(1/4)^(n-5)

Answer 3

a2a3/a1a2=a3/a1=q^2
a5/a2=q^3=1/8,
q=1/2,q^2=1/4,
a1=a2/q=2/(1/2)=4,
a1a2=4*2=8,
a1a2+a2a3+...+anan+1
=8[1-(1/4)^2n]/(1-1/4)
=32[1-(1/4)^2n]/3。如果我的答案对您有帮助，请点击下面的“采纳答案”按钮，送咱一朵小
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Answer 4

a2=2=a1q
a5=a1q^4=1/4
q^3=1/8
q=1/2
a1=4
所以An=4*1/2^n-1=2^3-n
anan+1=2^3-n*2^2-n=2^5-2n
所以
anan+1/an-1an=1/4
a1a2=4*2=8
所以
anan+1为以8为首项,1/4为公比的等比
所以a1a2+a2a3+```anan+1=(8-2^5-2n*1/4)/(1-1/4)=4(8-2^(3-2n))/3