已知递增数列{an}的前n项和为Sn,且满足a1=1,4Sn-4n+1=an2.设bn=1求详解,感激不尽.
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已知递增数列{an}的前n项和为Sn,且满足a1=1,4Sn-4n+1=an2.设bn=1求详解,感激不尽.
已知递增数列{an}的前n项和为Sn,且满足a1=1,4Sn-4n+1=an2.设bn=1
求详解,感激不尽.
已知递增数列{an}的前n项和为Sn,且满足a1=1,4Sn-4n+1=an2.设bn=1求详解,感激不尽.
1.
n≥2时,
4Sn-4n+1=an² (1)
4S(n-1)-4(n-1)+1=a(n-1)² (2)
(1)-(2)
4[Sn-S(n-1)]+4=an²-a(n-1)²
4an-4=an²-a(n-1)²
an²-4an+4=a(n-1)²
(an -2)²=a(n-1)²
|an-2|=|a(n-1)|
数列为递增数列,又a1=1>0,an≥1
|an-2|=a(n-1)
an-2=a(n-1)或an-2=-a(n-1)
an=a(n-1)+2或an=2-a(n-1)
an=2-a(n-1)时,a2=2-a1=2-1=1=a1,与已知数列是递增数列矛盾,舍去
因此只有an-2=a(n-1)
an-a(n-1)=2,为定值
数列{an}是以1为首项,2为公差的等差数列
2.
an=1+2(n-1)=2n-1
[am²+a(m+1)²-a(m+2)²]/[ama(m+1)]
=[(2m-1)²+(2m+1)²-(2m+3)²]/[(2m-1)(2m+1)]
=(4m²-12m-7)/[(2m-1)(2m+1)]
=(2m+1)(2m-7)/[(2m-1)(2m+1)]
=(2m-7)/(2m-1)
=(2m-1-6)/(2m-1)
=1- 6/(2m-1)
要[am²+a(m+1)²-a(m+2)²]/[ama(m+1)]为整数,6/(2n-1)为整数,6能被2n-1整除
6=1×6=2×3
令2m-1=1,解得m=1；
令2m-1=6,解得m=7/2,不是整数,舍去
令2m-1=2,解得m=3/2,不是整数,舍去
令2m-1=3,解得m=2
综上,得m=1或m=2
3.
bn=1/[ana(n+1)]=1/[(2n-1)(2n+1)]=(1/2)[1/(2n-1)-1/(2n+1)]
Tn=b1+b2+...+bn
=(1/2)[1/1-1/3+1/3-1/5+...+1/(2n-1)-1/(2n+1)]
=(1/2)[1- 1/(2n+1)]
=n/(2n+1)
λTn