已知数列an各项均为正数,其前n项和为Sn,且An²+2An=4Sn (1)求an通项公式 (2)
已知数列an各项均为正数,其前n项和为Sn,且An²+2An=4Sn(1)求an通项公式(2)若{Bn}=4/An²,Tn=b1+b2+...+bn,...
已知数列an各项均为正数,其前n项和为Sn,且An²+2An=4Sn
(1)求an通项公式
(2)若{Bn}=4/An²,Tn=b1+b2+...+bn,求证Tn<5/3 展开
(1)求an通项公式
(2)若{Bn}=4/An²,Tn=b1+b2+...+bn,求证Tn<5/3 展开
2个回答
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(1)当n=1时,a1=s1=
1
4
a 21
+
1
2
a1-
3
4
,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 21
+
1
2
a1-
3
4
,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
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