已知公差大于零的等差数列{an}的前n项为Sn,且满足a3*a4=117,a2+a5=22.
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已知公差大于零的等差数列{an}的前n项为Sn,且满足a3*a4=117,a2+a5=22.
1.求通项公式an;2.求Sn的最小值;3.若数列{bn}是等差数列,且bn=Sn/n+c,求非零常数c.
1.求通项公式an;2.求Sn的最小值;3.若数列{bn}是等差数列,且bn=Sn/n+c,求非零常数c.
1.
a2+a5=a3+a4=22
a3*a4=117
a3、a4是方程x^2-22x+117=0的根,
x1=9、x2=13
所以a3=9、a4=13(因a4>a3)
d=a4-a3=4
a1=a3-2d=9-8=1
an=1+4(n-1)=4n-3;
2.
an=1+4(n-1)=4n-3
Sn=4n(n+1)/2-3n=2n^2-n
因d>0,所以Sn单调递增,所以S1最大=a1=1;
3.
方法一:
bn=Sn/(n+c)
=(2n^2-n)/(n+c)
=n(2n-1)/(n+c)
b(n-1)=(n-1)(2n-3)/(n-1+c)
d=bn-b(n-1)
=n(2n-1)/(n+c)-(n-1)(2n-3)/(n-1+c)
=(2n^2-n)/(n+c)-(2n^2-5n+3)/(n-1+c)
(n+c)(n-1+c)d=(2n^2-n)(n-1+c)-(2n^2-5n+3)(n+c)
(n^2-n+2cn-c+c^2)d=2n^2+4cn-2n-3c
(d-2)n^2+(2c-1)(d-2)n+c(-d+dc+3)=0
要使上式恒成立,则
d-2=0,(2c-1)(d-2)=0,c(-d+dc+3)=0
d=2,(2c+1)c=0
所以d=2,c=-1/2 ;
方法二:
因等差数列可表示成kn+e的形式,k、e为常数,所以
bn=Sn/(n+c)
=(2n^2-n)/(n+c)
=kn+e
2n^2-n=(n+c)(kn+e)=kn^2+(ck+e)n+ce
(k-2)n^2+(ck+e+1)n+ce=0
要使上式恒成立,则
k-2=0,ck+e+1=0,ce=0
k=2,ck+e+1=0,e=0
k=2,2c+1=0,e=0
c=-1/2.
a2+a5=a3+a4=22
a3*a4=117
a3、a4是方程x^2-22x+117=0的根,
x1=9、x2=13
所以a3=9、a4=13(因a4>a3)
d=a4-a3=4
a1=a3-2d=9-8=1
an=1+4(n-1)=4n-3;
2.
an=1+4(n-1)=4n-3
Sn=4n(n+1)/2-3n=2n^2-n
因d>0,所以Sn单调递增,所以S1最大=a1=1;
3.
方法一:
bn=Sn/(n+c)
=(2n^2-n)/(n+c)
=n(2n-1)/(n+c)
b(n-1)=(n-1)(2n-3)/(n-1+c)
d=bn-b(n-1)
=n(2n-1)/(n+c)-(n-1)(2n-3)/(n-1+c)
=(2n^2-n)/(n+c)-(2n^2-5n+3)/(n-1+c)
(n+c)(n-1+c)d=(2n^2-n)(n-1+c)-(2n^2-5n+3)(n+c)
(n^2-n+2cn-c+c^2)d=2n^2+4cn-2n-3c
(d-2)n^2+(2c-1)(d-2)n+c(-d+dc+3)=0
要使上式恒成立,则
d-2=0,(2c-1)(d-2)=0,c(-d+dc+3)=0
d=2,(2c+1)c=0
所以d=2,c=-1/2 ;
方法二:
因等差数列可表示成kn+e的形式,k、e为常数,所以
bn=Sn/(n+c)
=(2n^2-n)/(n+c)
=kn+e
2n^2-n=(n+c)(kn+e)=kn^2+(ck+e)n+ce
(k-2)n^2+(ck+e+1)n+ce=0
要使上式恒成立,则
k-2=0,ck+e+1=0,ce=0
k=2,ck+e+1=0,e=0
k=2,2c+1=0,e=0
c=-1/2.
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