Question

已知等比数列{a_n}中,a₁+a₃=10,a₄+a₆=(n∈N^*).

(1)求数列{a_n}的通项公式;

(2)(理)试比较与2lg2的大小,并说明理由.

(文)试比较与lg2的大小,并说明理由.

Answer 1

解：(1)设数列{a_n}的公比为q,则根据条件得

即                                             

②÷①得q³=,所以q=.代入①解得a₁=8.                                      

所以a_n=a₁q^n-1=8·()^n-1=()^n-4.                                               

(2)(理)因为-2lg2                                  

=-2lg2

=-2lg2

=-2lg2                                           

=()lg-2lg2=lg2+lg2-2lg2

=lg2lg2=(-1)lg2,                                               

设g(n)=(-1)lg2,

因为g(n)是关于n的减函数,所以g(n)≤g(n)|_max=g(1)(n∈N^*),

即(-1)lg2≤［(-1)lg2］|_max=(-1)lg2=0.

所以≤2lg2.                                     

(文)因为                              

=

=

=                                           

=()lglg2

=lg2+lg2+lg2

=lg2＞0,                                                               

所以lg2.