星座魔法妹妹
幼苗
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证明:(1)若c=0,则a n =a 1 +(n-1)d, S n =
n[(n-1)d+2a]
2 , b n =
n S n
n 2 =
(n-1)d+2a
2 .
当b 1 ,b 2 ,b 4 成等比数列时,则 b 2 2 = b 1 b 4 ,
即: (a+
d
2 ) 2 =a(a+
3d
2 ) ,得:d 2 =2ad,又d≠0,故d=2a.
因此: S n = n 2 a , S nk =(nk ) 2 a= n 2 k 2 a , n 2 S k = n 2 k 2 a .
故: S nk = n 2 S k (k,n∈N*).
(2) b n =
n S n
n 2 +c =
n 2
(n-1)d+2a
2
n 2 +c
=
n 2
(n-1)d+2a
2 +c
(n-1)d+2a
2 -c
(n-1)d+2a
2
n 2 +c
=
(n-1)d+2a
2 -
c
(n-1)d+2a
2
n 2 +c .①
若{b n }是等差数列,则{b n }的通项公式是b n =A n +B型.
观察①式后一项,分子幂低于分母幂,
故有:
c
(n-1)d+2a
2
n 2 +c =0 ,即 c
(n-1)d+2a
2 =0 ,而
(n-1)d+2a
2 ≠0 ,
故c=0.
经检验,当c=0时{b n }是等差数列.
1年前
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