我不爱张sir
幼苗
共回答了18个问题采纳率:88.9% 举报
(Ⅰ)由已知a n =-6n-2,故{a n }是以a 1 =-8为首项公差为-6的等差数列.
所以S n =-3n 2 -5n.
(Ⅱ)因为c n =a n +8n+3=-6n-2+8n+3=2n+1(n∈N*), d n+1 = c d n =2d n +1,因此d n+1 +1=2(d n +1)(n∈N*).
由于d 1 =c 1 =3,
所以{d n +1}是首项为d 1 +1=4,公比为2的等比数列.
故d n +1=4×2 n-1 =2 n+1 ,所以d n =2 n+1 -1.
(Ⅲ)解法一: g(
d n +1
2 )=g( 2 n )= 2 n-1 g(2)+2g( 2 n-1 ) ,
则 b n =
2 n-1 g(2)+2g( 2 n-1 )
2 n+1 =
a
4 +
g( 2 n-1 )
2 n ,b n+1 =
a
4 +
g( 2 n )
2 n+1 . b n+1 - b n =
g( 2 n )
2 n+1 -
g( 2 n-1 )
2 n =
2 n-1 a+2g( 2 n-1 )
2 n+1 -
g( 2 n-1 )
2 n =
a
4 .
因为a为常数,则数列{b n }是等差数列.
解法二:因为g(x 1 x 2 )=x 1 g(x 2 )+x 2 g(x 1 )成立,且g(2)=a,
故 g(
d n +1
2 )=g( 2 n )= 2 n-1 g(2)+2g( 2 n-1 ) =2 n-1 g(2)+2[2 n-2 g(2)+2g(2 n-2 )]=2×2 n-1 g(2)+2 2 g(2 n-2 )=2×2 n-1 g(2)+2 2 [2 n-3 g(2)+2g(2 n-3 )]=3×2 n-1 g(2)+2 3 g(2 n-3 )═(n-1)×2 n-1 g(2)+2 n-1 g(2)=n•2 n-1 g(2)=an•2 n-1 ,
所以 b n =
g(
d n +1
2 )
d n +1 =
an• 2 n-1
2 n+1 =
a
4 n .
则 b n+1 - b n =
a
4 .
由已知a为常数,因此,数列{b n }是等差数列.
1年前
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