在数1和2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个数的乘机记为An,令an=log2An n为正整数.(1) 求数列{An}的前n项和Sn(2) 求Tn=tana2×tana4+tana4×tana6+...+tana2n×tana2n+2

Question

在数1和2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个数的乘机记为An,令an=log2An n为正整数.
(1) 求数列{An}的前n项和Sn
(2) 求Tn=tana2×tana4+tana4×tana6+...+tana2n×tana2n+2

Answer 1

An=(1×2)^[(n+2)/2]=2^[(n+2)/2]=2^(1+n/2)=2*2^(n/2)=2*(√2)^n
(1)Sn=A1+A2+……+An=2√2*[1-(√2)^n]/(1-√2)=2(2+√2)*[(√2)^n-1]
(2)an=log2 An=log2 2^(1+n/2)=1+n/2
tana2n=tan(n+1)
则
tana2n×tana(2n+2)=tan(n+1)tan(n+2)
考虑tan1=tan[(n+2)-(n+1)]=[tan(n+2)-tan(n+1)]/[1+tan(n+2)tan(n+1)]
解得
tan(n+2)tan(n+1)=[tan(n+2)-tan(n+1)-tan1]/tan1
=cot1*[tan(n+2)-tan(n+1)]-1
故
Tn=tana2×tana4+tana4×tana6+...+tana2n×tana2n+2
=tan2×tan3+tan3×tan4+...+tan（n+1）×tan（n+2）
=cot1*(tan3-tan2)-1+cot1*(tan4-tan3)-1+……+cot1*[tan(n+2)-tan(n+1)]-1
=cot1*[tan(n+2)-tan2]-n
=tan(n+2-2)*[1+tan(n+2)tan2]*cot1-n
=tann/tan1*[1+tan(n+2)tan2]-n