Question

已知n∈N^*，数列{a_n}和{b_n}满足a₁=b₁=6,a₂=4，数列{b_n}的前n项和为S_n且S_n+1=S_n+n+6,

(1)求证:{b_n-2}是等比数列；

(2)若数列{a_n+1-a_n}(n∈N^*)是公差为1的等差数列，试比较a_n与b_n的大小.

Answer 1

解：(1)证明:当n≥2时，S_n+1=S_n+n+6,                   ①

S_n=S_n-1+n+5,                                                      ②

①-②,得b_n+1=b_n+1.                                                         

∴b_n+1-2=(b_n-2)，即n≥2时{b_n-2}是等比数列.

又S₂=S₁+1+6,

∴b₂=-b₁+1+6-b₁=4.

∴b₂-2= (b₁-2)=2,即n∈N^*时{b_n-2}是等比数列.                                

(2)由(1)知b_n-2=(b₁-2)·()^n-1,即b_n=2+8·()ⁿ.                                

由已知a₂-a₁=-2,∴a_n+1-a_n=(a₂-a₁)+(n-1)·1=n-3.

n≥2时，

a_n=(a_n-a_n-1)+(a_n-1-a_n-2)+…+(a₃-a₂)+(a₂-a₁)+a₁=(n-4)+(n-5)+…+(-1)+(-2)+6=.

n=1也合适.

∴a_n=(n∈N^*).                                                   

设f(n)=a_n-b_n=n²-n+7-8·()ⁿ=(n-)²+-8·()ⁿ.                        

当n≥4时(n-)²+为n的增函数，-8·()ⁿ也为n的增函数，

∴当n≥4时有f(n)≥f(4)=，即a_n-b_n≥.                                      

又f(1)=f(2)=f(3)=0,∴对n∈N^*都有a_n≥b_n.