已知{an)为等比数列,其前n项和为Sn,且Sn=2^n+a若bn=(2n-1)an,求数列｛bn}的前n项和Tn,我会做,只是每次做的最后结果都不同,麻烦列一下,我看下我哪里错了?
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已知{an)为等比数列,其前n项和为Sn,且Sn=2^n+a若bn=(2n-1)an,求数列｛bn}的前n项和Tn,我会做,只是每次做的最后结果都不同,麻烦列一下,我看下我哪里错了?
已知{an)为等比数列,其前n项和为Sn,且Sn=2^n+a
若bn=(2n-1)an,求数列｛bn}的前n项和Tn,
我会做,只是每次做的最后结果都不同,麻烦列一下,我看下我哪里错了?
已知{an)为等比数列,其前n项和为Sn,且Sn=2^n+a若bn=(2n-1)an,求数列｛bn}的前n项和Tn,我会做,只是每次做的最后结果都不同,麻烦列一下,我看下我哪里错了?
1,求出an通项公式
an = Sn - Sn-1 = (2^n+a) - (2^n-1 +a) = 2^(n-1)
q = an/an-1 = 2^(n-1)/2^(n-2) = 2
由等比数列求和公式
Sn = (a1- an*q)/(1-q) = 2^n -a1
所以 -a1 = a
a1 = S1 = 2^1 + a = 2+a
所以 2+a = -(-a1) = -a
a = -1
a1 = 1
所以 an = 1* 2^(n-1) = 2^(n-1)
2
设{Cn} = {2n-1},Cn公差 d = 2
q Tn - Tn = (c1a2+c2a3+c3a4+...+ cnan+1) - (c1a1+c2a2+c3a3+...+ cnan)
= -c1a1 + (-d)a2 + (-d)a3+...+ (-d) an + cnan+1
= -1 -d * (a2- an *q)/(1 -q) + cnan+1
= -1 -2* (2^n -2) + (2n-1) 2^n
= (2n-1 -2) 2^n -1 -2*(-2)
= (2n-3) 2^n +3
即
2 Tn - Tn = (2n-3) 2^n +3
Tn = (2n-3) 2^n +3
Sn=2^n+a
S(n-1)=2^(n-1)+a
Sn-S(n-1)=2^n+a-2^(n-1)-a
an=2^n-2^(n-1)
=2*2^(n-1)-2^(n-1)
=2^(n-1)
bn=(2n-1)an
=(2n-1)2^(n-1)
=n2^n-2^(n-1)
Tn=b1+b2+......
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Sn=2^n+a
S(n-1)=2^(n-1)+a
Sn-S(n-1)=2^n+a-2^(n-1)-a
an=2^n-2^(n-1)
=2*2^(n-1)-2^(n-1)
=2^(n-1)
bn=(2n-1)an
=(2n-1)2^(n-1)
=n2^n-2^(n-1)
Tn=b1+b2+...........bn
=1*2^1-2^0+2*2^2-2^1+..............+n2^n-2^(n-1)
=1*2^1+2*2^2+.......n2^n -[2^0+2^1+.........+2^(n-1)]
令Mn=1*2^1+2*2^2+.......n2^n
2Mn=1*2^2+2*2^3+.......(n-1)2^n+n2^(n+1)
Mn-2Mn=-Mn
=1*2^1+1*2^2+.............+1*2^n-n2^(n+1)
=2(2^n-1)/(2-1)-n2^(n+1)
=2^(n+1)-2-n2^(n+1)
=2^(n+1)(1-n)-2
∴Mn=2^(n+1)(n-1)+2
2^0+2^1+........+2^(n-1)
=1*(2^n-1)/(2-1)
=2^n-1
∴Tn=Mn-Tn
=2^(n+1)(n+1)+2-2^n+1
=n2^(n+1)+2^(n+1)-2^n+3
=2n2^n-2^n+2*2^n+3
=(2n-1)2^n+2*2^n+3
=2^n(2n+1)+3
收起
Tn-2Tn=1+2.2^1+2.2^2+...+(2.2^n-1)-(2n-1)2^n
1-(2n-1)2^n+2(2^1+2^2+...+2^n-1）
你的这一步有问题，第一项好像没有了