已知{an}是等比数列，a2=2，a5=1⼀4，则a1*a2+a2*a3+……+an*(an+1)=

q³=a5/a2=1/8
q=1/2
a1=a2/q=16

令bn=ana(n+1)
=a1*q^(n-1)*a1*q^n
=a1²*q^(2n-1)
所以bn也是等比数列
b1=a1²*q=128
q'=q²=1/4
所以原式=128[1-(1/4)^n]/(1-1/4)]
=[512-2^(9-n)]/3

a2=2，a5=1/4

a5=a2*q^3

q=1/2
a1=4

T=a1*a2+a2*a3+……+an*(an+1)

q*T=q*a1*a2+q*a2*a3+....+q*an*a(n+1)
=a2^2+a3^2+....+a(n+1)^2

T/q=a1^2+a2^2+...+an^2

两式子相减有
q*T-T/q=a(n+1)^2-a1^2
a1=4,q=1/2,a(n+1)=a1*q^n=4*2^(-n)=2^(2-n)代入上式子有
(1/2)*T-T/(1/2)=2^(4-2n)-16

T=(1/3)*(32-2^(5-2n))

设a1=a,公比为q
a2=a*q=2
a5=a*q^4=1/4
q^3=1/8
q=1/2
a1=4

a1*a2+a2*a3+……+an*(an+1)
=q*(a1*a1+a2*a2+……+an*an)
=q*a^2*(1+q^2+q^4+……+q^2n)
=q*a^2*(1-q^(2n+2))/(1-q^2)
=1/2*4^2*(1-q^(2n+2))/(3/4)
=32/3*(1-1/2^(2n+2))）
=32/3*(1-1/4^(n+1))

答案是C 答案是C
答案是C 答案是C

设a1=a,公比为q
a2=a*q=2
a5=a*q^4=1/4
q^3=1/8
q=1/2
a1=4

a1*a2+a2*a3+……+an*(an+1)
=q*(a1*a1+a2*a2+……+an*an)
=q*a^2*(1+q^2+q^4+……+q^2n)
=q*a^2*(1-q^(2n+2))/(1-q^2)
=1/2*4^2*(1-q^(2n+2))/(3/4)
=32/3*(1-1/2^(2n+2))）
=32/3*(1-1/4^(n+1))

n是可以赋值的
当n=-n+1时。。。。你懂的