Question

已知数列{a_n}中,a₁=2,S_n为其前n项和,且2a_n,,na_n(n∈N^*)成等差数列.设b_n=,记数列{b_n}的前n项和为T_n.

(1)求数列{a_n}的通项公式;

(2)设T_n=P,试比较与2ⁿ的大小,并证明你的结论.

Answer 1

解:(1)依题意,得2·=2a_n+na_n,即3S_n=(2+n)a_n(n∈N^*),                                    ?

∴3S_n-1=(1+n)a_n-1(n≥2),以上两式相减,得?

3(S_n-S_n-1)=(2+n)a_n-(1+n)a_n-1,整理得3a_n=(2+n)a_n-(1+n)a_n-1,?

∴=.                                                                                         ?

由此得a_n=a₁····…··?

=a₁····…···?

=a₁·.?

又a₁=2,∴a_n=n(n+1)(n∈N^*).                                                                      ?

(2)T_n=b₁+b₂+…+b_n=++…+?

=1-+-+…+-?

=1-=,                                                                                             ?

∴T_n==1,即p=1,则?

==n(n+1).?

∴=a_n.                                                                                                        ?

当n=1时,a₁=1(1+1)=2,∴a₁=2¹;?

当n=2,3,4时,则a₂=2(2+1)=6,∴a₂＞2²,?

a₃=3(3+1)=12,∴a₃＞2³,a₄=4(4+1)=20,?

∴a₄＞2⁴;?

当n≥5时,猜想a_n＜2ⁿ.                                                                                             ?

证法一:∵2ⁿ=(1+1)ⁿ=C⁰_n+C¹_n+C²_n+…+C^n-2_n+C^n-1_n+Cⁿ_n?

≥2(C⁰_n+C¹_n+C²_n)=2［1+n+］?

=n²+n+2＞n²+n=n(n+1)=a_n,?

∴当n≥5时,a_n＜2ⁿ.?

综上所述,当n=1时,=2ⁿ;?

当n=2,3,4时,＞2ⁿ;?

当n≥5时,＜2ⁿ.                                                                                           ?

证法二:①当n=5时,a₅=5(5+1)=30＜2⁵成立;?

②假设当n=k(k≥5)时,猜想成立,则有a_k＜2^k,即k(k+1)＜2^k,那么,??

当n=k+1时,有a_k+1=(k+1)(k+2)=a_k+2k+2＜2^k+2k+2,?

又∵k≥5时,2^k-1=C⁰_k-1+C¹_k-1+C²_k-1+…=1+(k-1)+C²_k-1+…=k+C²_k-1+…＞k+1,?

∴2(k+1)＜2·2^k-1=2^k,故2^k+2k+2＜2^k+2^k=2^k+1,即a_k+1＜2^k+1成立.?

由①②可知,a_n＜2ⁿ对一切n≥5的正整数都成立.?

综上所述,当n=1时,=2ⁿ;当n=2,3,4时,＞2ⁿ;??

当n≥5时,＜2ⁿ.