Question

在數(shù)列{a_n},{b_n}中,a₁=2,b₁=4,且a_n,b_n,a_n+1成等差數(shù)列,b_n,a_n+1,b_n+1成等比數(shù)列(n∈N^*).

(1)求a₂,a₃,a₄及b₂,b₃,b₄,由此猜測{a_n},{b_n}的通項公式,并證明你的結(jié)論;

(2)證明++…+＜.

Answer 1

本題主要考查等差數(shù)列、等比數(shù)列、數(shù)學(xué)歸納法、不等式等基礎(chǔ)知識,考查綜合運用數(shù)學(xué)知識進(jìn)行歸納、總結(jié)、推理、論證等能力.

解:(1)由條件得2b_n=a_n+a_n+1,a_n+1²=b_nb_n+1.

由此可得a₂=6,b₂=9,a₃=12,b₃=16,a₄=20,b₄=25.                                     

猜測a_n=n(n+1),b_n=(n+1)².                                                       

用數(shù)學(xué)歸納法證明:

①當(dāng)n=1時,由上可得結(jié)論成立.

②假設(shè)當(dāng)n=k時,結(jié)論成立,即a_k=k(k+1),b_k=(k+1)²,

那么當(dāng)n=k+1時,a_k+1=2b_k-a_k=2(k+1)²-k(k+1)=(k+1)(k+2),

b_k+1==(k+2)².

所以當(dāng)n=k+1時,結(jié)論也成立.

由①②,可知a_n=n(n+1),b_n=(n+1)²對一切正整數(shù)都成立.                           

(2).

n≥2時,由(1)知a_n+b_n=(n+1)(2n+1)＞2(n+1)n.                                       

故++…+＜+［++…+］

=+(-+-+…+)

=+()＜+=.