作业帮已知等差数列{an}的首项为a,公差为b,且不等式ax^2-3x+2>0的解集为(-无穷,1)∪(b,+无穷)若数列{bn}满足bn=an*(2^n),求数列bn前n项和Tn.
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已知等差数列{an}的首项为a,公差为b,且不等式ax^2-3x+2>0的解集为(-无穷,1)∪(b,+无穷)若数列{bn}满足bn=an*(2^n),求数列bn前n项和Tn.
已知等差数列{an}的首项为a,公差为b,且不等式ax^2-3x+2>0的解集为(-无穷,1)∪(b,+无穷)若数列{bn}满足bn=an*(2^n),求数列bn前n项和Tn.
已知等差数列{an}的首项为a,公差为b,且不等式ax^2-3x+2>0的解集为(-无穷,1)∪(b,+无穷)若数列{bn}满足bn=an*(2^n),求数列bn前n项和Tn.
由ax^2-3x+2>0解集为(-∞,1)U(b,+∞)得x=1时,ax^2-3x+2=0
a-3+2=0
a=1
x^2-3x+2>0
(x-1)(x-2)>0
x2
b=2
数列{an}是以1为首项,2为公差的等差数列.
an=1+2(n-1)=2n-1
bn=an×2ⁿ=(2n-1)×2ⁿ
Tn=b1+b2+...+bn=1×2+3×2^2+5×2^3+...+(2n-1)×2ⁿ
2Tn=1×2^2+3×2^3+...+(2n-3)×2ⁿ+(2n-1)×2^(n+1)
Tn-2Tn=-Tn=2+2×2^2+2×2^3+...+2×2ⁿ-(2n-1)×2^(n+1)
=2×(2+2^2+...+2ⁿ) -(2n-1)×2^(n+1) -2
=2×2×(2ⁿ-1)/(2-1)-(2n-1)×2^(n+1) -2
=(3-2n)×2^(n+1) -6
Tn=(2n-3)×2^(n+1) +6
由韦达定理知 1+b=3/a 1*b=2/a 得a=1 b=2 即an=2n-1 bn=(2n-1)*2^n Tn=b1+b2+b3+b4+...+bn 2Tn=2b1+2b2+2b3+...+2bn 由2-1得
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由韦达定理知 1+b=3/a 1*b=2/a 得a=1 b=2 即an=2n-1 bn=(2n-1)*2^n Tn=b1+b2+b3+b4+...+bn 2Tn=2b1+2b2+2b3+...+2bn 由2-1得 Tn=-(2+2*2^2+2*2^3+...+2*2^n-(2n-1)*2^(n+1)) =-2(2^(n+1)-3-2n *2^(n+1)+2^(n+1)) =2^(n+3) *(n-1)+6
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