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
问题描述:
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
答
(1)a2-a1=-2,a3-a2=-1由{an+1-an}成等差数列知其公差为1,
故an+1-an=-2+(n-1)•1=n-3;
b2-b1=-2,b3-b2=-1,
由{bn+1-bn}等比数列知,其公比为12,故bn+1-bn=-2•(
12)n-1,(6分)
an=(an-an-1)+(an-1-an-2)+(an-2-an-3)+…+(a2-a1)+a1
=(n-1)•(-2)+
(n-1)(n-2)2•1+6
=n2-3n+22-2n+8
=n2-7n+182,(8分)
bn=(bn-bn-1)+(bn-1-bn-2)+(bn-2-bn-3)+…+(b2-b1)+b1
=-2(1-(
12)n-2)1-
12+6
=2+24-n.
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^...