已知公差大于零的等差数列{an}的前n项和Sn,且满足a3*a4=117,a2+a5=22
求:1.等差数列{an}
2.若数列{bn}是等差数列,bn=Sn/(n+c),求非零常数c;
3、f(n)=bn/[(n+36)bn+1](n∈N+)的最大值
人气:483 ℃ 时间:2019-08-20 05:20:49
解答
因为an是公差d>0的等差数列,
所以 a2+a5=22=a3+a4
a3*a4=117
所以解得a3=9,a4=13
所以公差d=a4-a3=13-9=4
所以a1=1
1)、an=a1+(n-1)*d=1+(n-1)*4=4n-3
2)、Sn=(a1+an)*n/2=(1+4n-3)*n/2=n(2n-1)
所以bn=n(2n-1)/(n+c)是等差数列,且c≠0
则n没有二次项,所以c=-0.5
3、bn=2n
f(n)=2n/〔(n+36)*2(n+1)〕=1/(n+37+36/n)≤1/(37+2√36)=1/7
即当n=36/n,得n=6时,f(n)max=f(6)=1/7
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