一道数列题,已知数列{an}中,a1=1,点P(An,An+1)在直线y=x+1上.数列{bn}是等比数列,tn=anb1+an-1b2+…+a2bn-1+a1bn,t1=1,t2=4,求tn.
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![一道数列题,已知数列{an}中,a1=1,点P(An,An+1)在直线y=x+1上.数列{bn}是等比数列,tn=anb1+an-1b2+…+a2bn-1+a1bn,t1=1,t2=4,求tn.](/uploads/image/z/812523-3-3.jpg?t=%E4%B8%80%E9%81%93%E6%95%B0%E5%88%97%E9%A2%98%2C%E5%B7%B2%E7%9F%A5%E6%95%B0%E5%88%97%7Ban%7D%E4%B8%AD%2Ca1%3D1%2C%E7%82%B9P%EF%BC%88An%2CAn%2B1%29%E5%9C%A8%E7%9B%B4%E7%BA%BFy%3Dx%2B1%E4%B8%8A.%E6%95%B0%E5%88%97%7Bbn%7D%E6%98%AF%E7%AD%89%E6%AF%94%E6%95%B0%E5%88%97%2Ctn%3Danb1%2Ban-1b2%2B%E2%80%A6%2Ba2bn-1%2Ba1bn%2Ct1%3D1%2Ct2%3D4%2C%E6%B1%82tn.)
一道数列题,已知数列{an}中,a1=1,点P(An,An+1)在直线y=x+1上.数列{bn}是等比数列,tn=anb1+an-1b2+…+a2bn-1+a1bn,t1=1,t2=4,求tn.
一道数列题,
已知数列{an}中,a1=1,点P(An,An+1)在直线y=x+1上.数列{bn}是等比数列,tn=anb1+an-1b2+…+a2bn-1+a1bn,t1=1,t2=4,求tn.
一道数列题,已知数列{an}中,a1=1,点P(An,An+1)在直线y=x+1上.数列{bn}是等比数列,tn=anb1+an-1b2+…+a2bn-1+a1bn,t1=1,t2=4,求tn.
等差数列{an}:a1=1;an=n;
t1=b1=1;t2=2b1+b2=4;
b1=1;b2=2;
故等比数列{bn}:b1=1;bn=2^(n-1);
tn=nb1+an-1b2+.+a2bn-1+a1bn
=n+(n-1)*2+(n-2)*2^2+(n-3)*2^3+...+(n-(n-2))*2^(n-2)+(n-(n-1))*2^(n-1)
=n*(1+2+2^2+2^3+...+2^(n-2)+2^(n-1))-(1*2+2*2^2+3*2^3+...+(n-2)*2^(n-2)+(n-1)*2^(n-1))
=n*(2^n-1)-(n-2)*2^n-2
=2^(n+1)-n-2
减号后面的等差等比混合式子的计算如下:
令 x=1*2+2*2^2+3*2^3+...+(n-2)*2^(n-2)+(n-1)*2^(n-1) (1)
则2x= 1*2^2+2*2^3+...+(n-(n-3)*2^(n-2)+(n-(n-2))*2^(n-1)+(n-1)*2^n (2)
(2)-(1)得x=(n-1)*2^n-2^n+2=(n-2)*2^n+2
点P(An,An+1)在直线y=x+1上
即:An+1=An+1
a1=1
故,An=n,(n≥1)
数列{bn}是等比数列,
不妨设公比为q,
则:bn=b1*q^(n-1)
tn=anb1+an-1b2+…+a2bn-1+a1bn
=n*b1+(n-1)*b1*q+…+2*b1*q^(n-2)+b1*q^(n-1)
=b1...
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点P(An,An+1)在直线y=x+1上
即:An+1=An+1
a1=1
故,An=n,(n≥1)
数列{bn}是等比数列,
不妨设公比为q,
则:bn=b1*q^(n-1)
tn=anb1+an-1b2+…+a2bn-1+a1bn
=n*b1+(n-1)*b1*q+…+2*b1*q^(n-2)+b1*q^(n-1)
=b1*[n+(n-1)*q+…+2*q^(n-2)+q^(n-1)]
t1=b1*a1=b1=1
t2=b1*[2+q]=4
解得:b1=1,q=2
即:bn=2^(n-1)
tn=b1*[n+(n-1)*q+…+2*q^(n-2)+q^(n-1)]
=n+(n-1)*2+…+2*2^(n-2)+2^(n-1)
=希格玛(i*2^(n-i))i=1到n。
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