已知{an}是各项均为整数的等比数列且a1+a2=2(1⼀a1+1⼀a2),a3+a4+a5=64(1⼀a3+1⼀a4+1⼀a5)

1。求{an}通项公式2。设bn=(an+1/an)^2,求数列{bn}的前n项和Sn
2025-04-08 11:22:16
推荐回答(1个)
回答1:

a1+a1q=2(1/a1+1/a1q)
a1(1+q)=2(1+q)/a1q
a1^2q=2 (1)
a1(q^2+q^3+q^4)=64(q^2+q+1)/a1q^4
a1^2q^6=64 (2)
(2)/(1)
q^5=32 q=2 a1=1
an=2^(n-1)
2. bn=(an+1/an)^2=4^(n-1)+2+1/4^(n-1)
Sn=(1-4^n)/(1-4)+2n+[1-(1/4)^n]/(1-1/4)
=(4^n-1)/3+2n+4[1-(1/4)^n]/3