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{C
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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{C
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
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{C
∵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^(3-n)]+[2n+2n(n-1)/2]
∴设 Tn=[1*2^2 +2*2^1 +3*2^0 +...(n-1)*2^(4-n) +n*2^(3-n)].[1]
2Tn=[1*2^3 +2*2^2 +3*2^1 +4*2^0 +...+(n-1)*2^(5-n)+n*2^(4-n)].[2]
[2]-[1]=Tn=[2^3+2^2+2^1+2^0+...+2^(4-n)]-n*2^(3-n)= 8(1-2^n)/1-2 - n*2^(3-n)
∴Sn=8(1-2^n)/1-2 - n*2^(3-n)) + [2n+2n(n-1)/2]
(3)Cn=an*(b(n+2)-2)=(8-2n) * 2^(1-n)
∴当n取得最小1时,8-2n=6为最大且2^(1-n)在n≥1中为单调减函数,即n=1时2^(1-n)=1为最大
∴Cn(MAX)=6
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(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-a...
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(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.
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