已知数列{a n }的前n项和S n 满足 S n = 1 4 ( a n +1 ) 2 ,且a n >0.(1)求数

2025-06-22 11:35:55
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回答1:

(1)当n=1时,有 a 1 = S 1 =
1
4
( a 1 +1 ) 2
,∴a 1 =1
当n=2时,有 a 1 + a 2 =
1
4
( a 2 +1 ) 2
,∴a 1 =3
当n≥2时,有 a n = S n - S n-1 =
1
4
[( a n +1 ) 2 -( a n-1 +1 ) 2 ]

∴(a n +a n-1 )(a n -a n-1 -2)=0又∵a n >0,∴a n -a n-1 =2,
∴数列{a n }是以1为首项,以2为公差的等差数列,
∴a n =1+(n-1)×2=2n-1
(2)由于b n =20-a n =21-2n,则b 1 =19,b n -b n-1 =-2<0.
∴{b n }是递减数列,
b n =21-2n>0  
b n+1 =19-2n<0

∴n=10,即数列{b n }的前10项和最大.