已知数列{an}的前n项和为sn,且sn=2n^2+n,n∈N*,数列{bn}满足an=4log2（bn）+3,n∈N*1 求an,bn2 求数列{an·bn}的前N项和为Tn
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已知数列{an}的前n项和为sn,且sn=2n^2+n,n∈N*,数列{bn}满足an=4log2（bn）+3,n∈N*1 求an,bn2 求数列{an·bn}的前N项和为Tn
已知数列{an}的前n项和为sn,且sn=2n^2+n,n∈N*,数列{bn}满足an=4log2（bn）+3,n∈N*
1 求an,bn
2 求数列{an·bn}的前N项和为Tn
已知数列{an}的前n项和为sn,且sn=2n^2+n,n∈N*,数列{bn}满足an=4log2（bn）+3,n∈N*1 求an,bn2 求数列{an·bn}的前N项和为Tn
(1)
Sn=2n^2+n
S(n-1)=2(n-1)^2+n-1
=2n^2-4n+2+n-1
=2n^2-3n+1
Sn-S(n-1)
=an
=2n^2+n-2n^2+3n-1
=4n-1
an=4log2 (bn)+3=4n-1
4log2(bn)=4n-4
log2(bn)=n-1
bn=2^(n-1)
(2)
an*bn=(4n-1)2^(n-1)=4n*2^(n-1)-2^(n-1)
=n2^(n-1+2)-2^(n-1)
=n2^(n+1)-2^(n-1)
Tn=1*2^2-2^0+2*2^3-2^1+.+n2^(n+1)-2^(n-1)
=1*2^2+2*2^3+.+n2^(n+1)-(2^0+2^1+.+2^(n-1))
∵2^0+2^1+.+2^(n-1)是首项k1=1,公比q=2的等比数列
2^0+2^1+.+2^(n-1)
=1*(2^n-1)/(2-1)
=2^n-1
设
Mn=1*2^2+2*2^3+.+n2^(n+1)
2Mn=2*2^2+2*2^4+.+n2^(n+2)
两式相减得
2Mn-Mn=Mn
=1*2^3+2*2^4+.+n2^(n+2)-1*2^2-2*2^3-.-n2^(n+1)
=-2^2-2^3-.-2^(n+1)+n2^(n+2)
∵数列-2^2-2^3-.-2^(n+1)是首项=-4,公比q=2的等比数列
-2^2-2^3-.-2^(n+1)
=-4(2^n-1)(2-1)
=-4(2^n-1)
所以
Mn=n2^(n+2)-4(2^n-1)
=n2^(n+2)-2^(n+2)+4
=(n-1)2^(n+2)+4
(1)
Sn=2n^2+n (1)
S(n-1) =2(n-1)^2+(n-1) (2)
(1)-(2)
an=4n-1
an= 4log(2)bn + 3
4n-1 = 4log(2)bn + 3
log(2)bn = n-1
bn = 2^(n-1...
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(1)
Sn=2n^2+n (1)
S(n-1) =2(n-1)^2+(n-1) (2)
(1)-(2)
an=4n-1
an= 4log(2)bn + 3
4n-1 = 4log(2)bn + 3
log(2)bn = n-1
bn = 2^(n-1)
(2)
cn=an.bn
= (4n-1) . 2^(n-1)
= 4(n. 2^(n-1) ) - 2^(n-1)
consider
1+x+x^2+...+x^n = (x^(n+1) -1)/(x-1)
1+2x+..+nx^(n-1) =[(x^(n+1) -1)/(x-1)]'
= [nx^(n+1)- (n+1)x^n +1]/(x-1)^2
put x=2
1.2^0+2.2^1+...+n.2^(n-1) = n. 2^(n+1)-(n+1)2^n + 1
Tn = c1+c2+..+cn
= 4[n. 2^(n+1)-(n+1)2^n + 1] - (1/2)(2^n-1)
= 2^n . [ 8n-4(n+1)-1/2 ] + 9/2
= (4n-9/2). 2^n + 9/2
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