高三一轮数列 数列求和错位相减家庭作业高手进啊已知等差数列an的公差d大于0,且a2,a5是方程x^2-12x+27=0的两根,数列bn的前n项和为Tn,且Tn=1-1/2bn（1）求{an} {bn} 会求求出来是an=2n-1,bn=2/3*[(1/3)的n-1

高三一轮数列 数列求和错位相减家庭作业高手进啊已知等差数列an的公差d大于0,且a2,a5是方程x^2-12x+27=0的两根,数列bn的前n项和为Tn,且Tn=1-1/2bn（1）求{an} {bn} 会求求出来是an=2n-1,bn=2/3*[(1/3)的n-1
高三一轮数列 数列求和错位相减家庭作业高手进啊
已知等差数列an的公差d大于0,且a2,a5是方程x^2-12x+27=0的两根,数列bn的前n项和为Tn,且Tn=1-1/2bn
（1）求{an} {bn} 会求求出来是an=2n-1,bn=2/3*[(1/3)的n-1次方]
(2)记cn=anbn,求{cn}的前n项和Sn
就是求第二问的错位相减!

高三一轮数列 数列求和错位相减家庭作业高手进啊已知等差数列an的公差d大于0,且a2,a5是方程x^2-12x+27=0的两根,数列bn的前n项和为Tn,且Tn=1-1/2bn（1）求{an} {bn} 会求求出来是an=2n-1,bn=2/3*[(1/3)的n-1
an=2n-1,bn=2/3*[(1/3)^(n-1)]
∵cn=anbn
∴cn=2/3*(2n-1)(1/3)^(n-1)
∴Sn=2/3(1*1+3*1/3+5*(1/3)^2+……+(2n-3)*(1/3)^(n-2)+(2n-1)(1/3)^(n-1))
1/3Sn=2/3(1*1/3+3*(1/3)^2+5(1/3)^3+……+(2n-3)(1/3)^(n-1)+(2n-1)*(1/3)^n)
两式相减得到
2/3Sn=2/3(1+2*1/3+2*(1/3)^2+……+2*(1/3)^(n-1)-(2n-1)*(1/3)^n)
∴Sn=2(1+1/3+(1/3)^2+……+(1/3)^(n-1))-(2n-1)(1/3)^n-1
=3(1-(1/3)^n)-(2n-1)(1/3)^n-1
=2-(2n-1+3)(1/3)^n
=2-2(n+1)(1/3)^n

x^2-12x+27=0
(x-3)(x-9)=0
x=3或x=9
公差d大于0
所以a2=3,a5=9
a5=a2+3d
9=3+3d
d=2
a2=a1+d
3=a1+2
a1=1
an=a1+(n-1)d
=1+2(n-1)
=2n-1
t1=1-1/2b1
b1=1-1/...

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x^2-12x+27=0
(x-3)(x-9)=0
x=3或x=9
公差d大于0
所以a2=3,a5=9
a5=a2+3d
9=3+3d
d=2
a2=a1+d
3=a1+2
a1=1
an=a1+(n-1)d
=1+2(n-1)
=2n-1
t1=1-1/2b1
b1=1-1/2b1
3b1/2=1
b1=2/3
tn=1-1/2bn
t(n-1)=1-1/2b(n-1)
两式相减得
bn=-1/2bn+1/2b(n-1)
3bn/2=1/2b(n-1)
3bn=b(n-1)
bn/b(n-1)=1/3
所以bn是以1/3为公比的等比数列
bn=b1q^(n-1)
=2/3*(1/3)^(n-1)
=2*(1/3)^n
cn=anbn
=(2n-1)*2*(1/3)^n
=2(2n-1)*(1/3)^n
sn=2*(2*1-1)*(1/3)+2*(2*2-1)*(1/3)^2+............+2(2n-1)*(1/3)^n
sn/2=(2*1-1)*(1/3)+(2*2-1)*(1/3)^2+............+(2n-1)*(1/3)^n
sn/6=(2*1-1)*(1/3)^2+(2*2-1)*(1/3)^3+............+(2n-1)*(1/3)^(n+1)
sn/2-sn/6=1/3+2/3^2+2/3^3+........+2/3^n-(2n-1)*(1/3)^(n+1)
sn/3=2/3+2/3^2+2/3^3+........+2/3^n-(2n-1)*(1/3)^(n+1)-1/3
sn/3=2/3*[1-(1/3)^n]/(1-1/3)-(2n-1)*(1/3)^(n+1)-1/3
sn/3=1-(1/3)^n-(2n-1)*(1/3)^(n+1)-1/3
sn/3=2/3-(1/3)^n-(2n-1)*(1/3)^(n+1)
sn=2-3*(1/3)^n-3*(2n-1)*(1/3)^(n+1)
sn=2-3*(1/3)^n-(2n-1)*(1/3)^n
sn=2-[3+(2n-1)]*(1/3)^n
sn=2-2(n+1)/3^n

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