Question

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

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

(2) 求證:++…+<.

Answer 1

 (1) 由題意得2bn=an+an+1,=bnbn+1,由此可得

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

猜測(cè)an=n(n+1),bn=(n+1)2.

用數(shù)學(xué)歸納法證明:

①當(dāng)n=1時(shí),由已知可得結(jié)論成立.

②假設(shè)當(dāng)n=k時(shí)結(jié)論成立,即

ak=k(k+1),bk=(k+1)2.

那么當(dāng)n=k+1時(shí),

ak+1=2bk-ak=2(k+1)2-k(k+1)=(k+1)(k+2),

bk+1==(k+2)2.

所以當(dāng)n=k+1時(shí),結(jié)論也成立.

由①②可知an=n(n+1),bn=(n+1)2對(duì)一切正整數(shù)都成立.

(2) 當(dāng)n=1時(shí),=<.

當(dāng)n≥2時(shí),由(1)知an+bn=(n+1)(2n+1)>2(n+1)n.

故++…+

<+×

=+×

=+×<+=,

綜上,原不等式成立.