Question

已知a1=1,a2=4,an+2=4an+1+an,bn=,n∈N+.

(1)求b1,b2,b3的值.

(2)設(shè)cn=bnbn+1,Sn為數(shù)列{cn}的前n項和,求證: Sn≥17n.

(3)求證:|b2n-bn|<·.

 

Answer 1

(1) (2) (3)見解析

【解析】(1)因為a1=1,a2=4,a3=4a2+a1=17,a4=72,所以b1=4,b2=,b3=.

(2)由an+2=4an+1+an

得=4+,

即bn+1=4+.

所以當(dāng)n≥2時,bn>4,于是c1=b1b2=17,cn=bnbn+1=4bn+1>17(n≥2),

所以Sn=c1+c2+…+cn≥17n.

(3)當(dāng)n=1時,結(jié)論|b2-b1|=<成立.

當(dāng)n≥2時,有|bn+1-bn|=|4+-4-|=

||≤|bn-bn-1|≤|bn-1-bn-2|≤…

≤|b2-b1|=.

所以|b2n-bn|≤|bn+1-bn|+|bn+2-bn+1|+…+|b2n-b2n-1|≤×[()n-1+()n+…+()2n-2]

=·<·(n≥2).

因此|b2n-bn|<·(n∈N+).