题目内容
已知数列{an}的各项均为正数,Sn是数列{an}的前n项和,且4Sn=an2+2an-3.(1)求数列{an}的通项公式;
(2)已知bn=2n,求Tn=a1b1+a2b2+…+anbn的值.
分析:(1)由题意知a1=s1=
+
a1-
,解得a1=3,由此能够推出数列{an}是以3为首项,2为公差的等差数列,所以an=3+2(n-1)=2n+1.
(2)由题意知Tn=3×21+5×22+…+(2n+1)•2n,2Tn=3×22+5×23+(2n-1)•2n+(2n+1)2n+1,二者相减可得到Tn=a1b1+a2b2+…+anbn的值.
1 |
4 |
a | 2 1 |
1 |
2 |
3 |
4 |
(2)由题意知Tn=3×21+5×22+…+(2n+1)•2n,2Tn=3×22+5×23+(2n-1)•2n+(2n+1)2n+1,二者相减可得到Tn=a1b1+a2b2+…+anbn的值.
解答:解:(1)当n=1时,a1=s1=
+
a1-
,解出a1=3,
又4Sn=an2+2an-3①
当n≥2时4sn-1=an-12+2an-1-3②
①-②4an=an2-an-12+2(an-an-1),即an2-an-12-2(an+an-1)=0,
∴(an+an-1)(an-an-1-2)=0,
∵an+an-1>0∴an-an-1=2(n≥2),
∴数列{an}是以3为首项,2为公差的等差数列,∴an=3+2(n-1)=2n+1.
(2)Tn=3×21+5×22+…+(2n+1)•2n③
又2Tn=3×22+5×23+(2n-1)•2n+(2n+1)2n+1④
④-③Tn=-3×21-2(22+23++2n)+(2n+1)2n+1-6+8-2•2n-1+(2n+1)•2n+1=(2n-1)•2n+2
1 |
4 |
a | 2 1 |
1 |
2 |
3 |
4 |
又4Sn=an2+2an-3①
当n≥2时4sn-1=an-12+2an-1-3②
①-②4an=an2-an-12+2(an-an-1),即an2-an-12-2(an+an-1)=0,
∴(an+an-1)(an-an-1-2)=0,
∵an+an-1>0∴an-an-1=2(n≥2),
∴数列{an}是以3为首项,2为公差的等差数列,∴an=3+2(n-1)=2n+1.
(2)Tn=3×21+5×22+…+(2n+1)•2n③
又2Tn=3×22+5×23+(2n-1)•2n+(2n+1)2n+1④
④-③Tn=-3×21-2(22+23++2n)+(2n+1)2n+1-6+8-2•2n-1+(2n+1)•2n+1=(2n-1)•2n+2
点评:本题考查数列的性质和应用,解题时要认真审题,仔细解答.

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