题目内容
已知Sn是数列{an}的前n项和,an>0,Sn=
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2 |
(Ⅰ)求Sn;
(Ⅱ)若数列{bn}满足b1=2,bn+1=2an+bn,求bn.
分析:(1)先利用S1=a1求得a1,再利用an=Sn-Sn-1求得当n≥2时an-an-1=1,判断出{an}是a1=1,公差为1的等差数列,进而利用等差数列的性质求得数列的前n项的和.
(2)把(1)中求得的an代入bn+1=2an+bn整理可求得bn+1-bn=2n,进而用叠加法求得数列{bn}的通项公式.
(2)把(1)中求得的an代入bn+1=2an+bn整理可求得bn+1-bn=2n,进而用叠加法求得数列{bn}的通项公式.
解答:解:(Ⅰ)当n=1时,S1=a1=
∴a12-a1=0.由a1>0,
得a1=1,当n≥2时,an=Sn-Sn-1=
-
整理得an2-an-12-an-an-1=0,?(an+an-1)(an-an-1-1)=0
由an>0,只有an-an-1-1=0,即an-an-1=1
所以{an}是a1=1,公差为1的等差数列,an=n, Sn=
;
(Ⅱ)由(Ⅰ)得bn+1-bn=2an=2n
所以(b2-b1)+(b3-b2)+(b4-b3)++(bn-bn-1)=21+22+23++2n-1
即bn-b1=2n-2,又b1=2,所以bn=2n.
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∴a12-a1=0.由a1>0,
得a1=1,当n≥2时,an=Sn-Sn-1=
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2 |
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整理得an2-an-12-an-an-1=0,?(an+an-1)(an-an-1-1)=0
由an>0,只有an-an-1-1=0,即an-an-1=1
所以{an}是a1=1,公差为1的等差数列,an=n, Sn=
n2+n |
2 |
(Ⅱ)由(Ⅰ)得bn+1-bn=2an=2n
所以(b2-b1)+(b3-b2)+(b4-b3)++(bn-bn-1)=21+22+23++2n-1
即bn-b1=2n-2,又b1=2,所以bn=2n.
点评:本题主要考查了等差数列的前n项的和.解题的关键是利用an=Sn-Sn-1得出数列的递推式.
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