在1与100之间插入n个实数,使得这n+2个数构成递增的等比数列,将这n+2个数的乘积记作Tn,令an=lg(Tn)(n
2个回答
(1)
设构成等比数列{an},公比为q
1为数列的第1项,100为数列的第n+2项.
100/1=q^(n+1)
q^(n+1)=100
Tn=a1a2...a(n+2)
=a1×(a1q)×...×[a1q^(n+1)]
=[a1^(n+2)]×q^[1+2+...+(n+1)]
=[1^(n+2)]×q^[(n+1)(n+2)/2]
=[q^[(n+1)/2]]^(n+2)
=[100^(1//2)]^(n+2)
=10^(n+2)
an=lg(Tn)=lg[10^(n+2)]=n+2
数列{an}的通项公式为an=n+2.
(2)
bn=tan(an) × tan[a(n+1)]
[tan(a(n+1) -t(an)]/[1+tan(an)×tan(a(n+1)]=tan[a(n+1)-an]=tan(n+1+2-n-2)=tan1
tan(an)×tan[a(n+1)]=[tan(a(n+1)-t(an)]/tan1 -1
Sn=b1+b2+...+bn
=[tan(a2)-tan(a1)+tan(a3)-tan(a2)+...+tan[a(n+1)-t(an)]]/tan1 -n
=[tan(a(n+1)) -tan(a1)]/tan1 -n
=[tan(n+3) -tan3]/tan1 -n
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