Question

已知數(shù)列{a_n}中,a₁=2,S_n為其前n項和,且2a_n,,na_n(n∈N^*)成等差數(shù)列.設(shè)b_n=,記數(shù)列{b_n}的前n項和為T_n.

(1)求數(shù)列{a_n}的通項公式;

(2)設(shè)T_n=P,試比較與2ⁿ的大小,并證明你的結(jié)論.

Answer 1

解:(1)依題意,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),以上兩式相減,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,則?

==n(n+1).?

∴=a_n.                                                                                                        ?

當(dāng)n=1時,a₁=1(1+1)=2,∴a₁=2¹;?

當(dāng)n=2,3,4時,則a₂=2(2+1)=6,∴a₂＞2²,?

a₃=3(3+1)=12,∴a₃＞2³,a₄=4(4+1)=20,?

∴a₄＞2⁴;?

當(dāng)n≥5時,猜想a_n＜2ⁿ.                                                                                             ?

證法一:∵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,?

∴當(dāng)n≥5時,a_n＜2ⁿ.?

綜上所述,當(dāng)n=1時,=2ⁿ;?

當(dāng)n=2,3,4時,＞2ⁿ;?

當(dāng)n≥5時,＜2ⁿ.                                                                                           ?

證法二:①當(dāng)n=5時,a₅=5(5+1)=30＜2⁵成立;?

②假設(shè)當(dāng)n=k(k≥5)時,猜想成立,則有a_k＜2^k,即k(k+1)＜2^k,那么,??

當(dāng)n=k+1時,有a_k+1=(k+1)(k+2)=a_k+2k+2＜2^k+2k+2,?

又∵k≥5時,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對一切n≥5的正整數(shù)都成立.?

綜上所述,當(dāng)n=1時,=2ⁿ;當(dāng)n=2,3,4時,＞2ⁿ;??

當(dāng)n≥5時,＜2ⁿ.