【人教版】2020版高考数学理科一轮复习:课时作业
由①-②得(an+an-1)(an-an-1-4)=0(an>0),则an-an-1=4,∴{an}是以4为公差的等差数列,即an=4n-2.
9 (2)b=1a+1+…+1+1n n+a1an+a2an+anan+an+1=111114n+4n+4+4n+8+…+4n+4?n-1?+4n+4n =1×11114n+[n+1+…+n+?n-1?+n+n
]
<1?11111?1?1n4×???n+n+1+n+1+…+n+1+n+1???=4×???n+?n+1???. 设f(n)=1n
n+n+1
,则f(n+1)-f(n)<0,
所以{f(n)}递减,1?4×?1?+n??133?nn+1??
≤4f(1)=8,即bn≤8.
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