已知{an}是等比数列,a1=1/2,a4=4,则a1a2+a2a3+……+ana(n+1)=?急！速答！
3个回答
展开全部
a4=a1q^3
4=1/2q^3
q^3=8
q=2
a1a2+a2a3+……+ana(n+1)
=a1*a1q+a1q*a1q^2+............+a1q^(n-1)*a1q^n
=(a1)^2*[1*q+q*q^2+q^2*q^3+........+q^(n-1)q^n]
=(a1)^2*[q+q^3+q^5+.......+q^(2n-1)]
=(a1)^2*{q*[1-(q^2)^n]/(1-q^2)}
=(a1)^2*{q*[1-q^2n]/(1-q^2)}
=(1/2)^2*{2*[1-2^2n]/(1-2^2)
=(1/2)^2*{2*[2^2n-1]/3
=(2)^-2*{2^(2n+1)-2]/3
=(2)^-2*2^(2n+1)/3-(2)^-2*2/3
=2^(2n-1)/3-1/6
4=1/2q^3
q^3=8
q=2
a1a2+a2a3+……+ana(n+1)
=a1*a1q+a1q*a1q^2+............+a1q^(n-1)*a1q^n
=(a1)^2*[1*q+q*q^2+q^2*q^3+........+q^(n-1)q^n]
=(a1)^2*[q+q^3+q^5+.......+q^(2n-1)]
=(a1)^2*{q*[1-(q^2)^n]/(1-q^2)}
=(a1)^2*{q*[1-q^2n]/(1-q^2)}
=(1/2)^2*{2*[1-2^2n]/(1-2^2)
=(1/2)^2*{2*[2^2n-1]/3
=(2)^-2*{2^(2n+1)-2]/3
=(2)^-2*2^(2n+1)/3-(2)^-2*2/3
=2^(2n-1)/3-1/6
本回答由提问者推荐
已赞过
已踩过<
评论
收起
你对这个回答的评价是？
展开全部
a1 = 1/2
a4 = a1q³ = 8
q = 2
an =2ⁿ-²
ana(n+1) = 2²ⁿ-³
a1a2 = 1/2 q' =4
a1a2 + a2a3 + … + ana(n+1) = (1/2)(4ⁿ - 1)/(4 - 1) = (4ⁿ - 1)/6
a4 = a1q³ = 8
q = 2
an =2ⁿ-²
ana(n+1) = 2²ⁿ-³
a1a2 = 1/2 q' =4
a1a2 + a2a3 + … + ana(n+1) = (1/2)(4ⁿ - 1)/(4 - 1) = (4ⁿ - 1)/6
已赞过
已踩过<
评论
收起
你对这个回答的评价是？
更多回答（1）
推荐律师服务：
若未解决您的问题，请您详细描述您的问题，通过百度律临进行免费专业咨询
