设数列{an}和{bn}满足a1=b1=6,a2=b2=4,a3=b3=3 ,且数列{an+1-an}是等差数列
设数列{an}和{bn}满足a1=b1=6,a2=b2=4,a3=b3=3 ,且数列{a(n+1)-an}是等差数列,{bn-2}是等比数列
(2)设{nbn}的前n项和为Sn,求Sn的表达式
(3)数列{Cn}满足Cn=an*(b(n+2)-2),求数列{Cn}的最大项
人气:401 ℃ 时间:2019-08-19 19:44:28
解答
∵数列{a(n+1)-an}是等差数列∴a2-a1=d=-2∴an=6-2(n-1)=8-2n∵{bn-2}是等比数列∴q=b2 -2/b1 -2=1/2∴bn-2=4乘以1/2^(n-1)∴bn=2^(3-n) +2∵nbn=n*2^(3-n) +2n∴Sn=b1+b2+b3+...+bn=[1*2^2 +2*2^1 +3*2^0 +...+n*2^...
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