等差数列﹛an﹜的各项均为正数,a1=3,前n项和为sn,﹛bn﹜为等比数列,b1=2,且b2s2=32,b3s3=120(1)求an与bn(2)求数列﹛anbn﹜的前n项和Tn
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![等差数列﹛an﹜的各项均为正数,a1=3,前n项和为sn,﹛bn﹜为等比数列,b1=2,且b2s2=32,b3s3=120(1)求an与bn(2)求数列﹛anbn﹜的前n项和Tn](/uploads/image/z/3623226-42-6.jpg?t=%E7%AD%89%E5%B7%AE%E6%95%B0%E5%88%97%EF%B9%9Ban%EF%B9%9C%E7%9A%84%E5%90%84%E9%A1%B9%E5%9D%87%E4%B8%BA%E6%AD%A3%E6%95%B0%2Ca1%3D3%2C%E5%89%8Dn%E9%A1%B9%E5%92%8C%E4%B8%BAsn%2C%EF%B9%9Bbn%EF%B9%9C%E4%B8%BA%E7%AD%89%E6%AF%94%E6%95%B0%E5%88%97%2Cb1%3D2%2C%E4%B8%94b2s2%3D32%2Cb3s3%3D120%EF%BC%881%EF%BC%89%E6%B1%82an%E4%B8%8Ebn%EF%BC%882%EF%BC%89%E6%B1%82%E6%95%B0%E5%88%97%EF%B9%9Banbn%EF%B9%9C%E7%9A%84%E5%89%8Dn%E9%A1%B9%E5%92%8CTn)
等差数列﹛an﹜的各项均为正数,a1=3,前n项和为sn,﹛bn﹜为等比数列,b1=2,且b2s2=32,b3s3=120(1)求an与bn(2)求数列﹛anbn﹜的前n项和Tn
等差数列﹛an﹜的各项均为正数,a1=3,前n项和为sn,﹛bn﹜为等比数列,b1=2,且b2s2=32,b3s3=120
(1)求an与bn
(2)求数列﹛anbn﹜的前n项和Tn
等差数列﹛an﹜的各项均为正数,a1=3,前n项和为sn,﹛bn﹜为等比数列,b1=2,且b2s2=32,b3s3=120(1)求an与bn(2)求数列﹛anbn﹜的前n项和Tn
(1)S2=a1+a2=a1+a1+d=6+d S3=3/2(a1+a3)=3a2=3a1+3d=9+3d
b2S2=(b1q)(6+d)=2q(6+d)=32 b3S3=(b1q^2)(9+3d)=(2q^2)(9+3d)=120
12q+2qd=32 18q^2+6q^2*d=120
6q+qd=16 3q^2+q^2*d=20
qd=16-6q
3q^2+(16-6q)q=20
3q^2+16q-6q^2-20=0
-3q^2+16q-20=0
3q^2-16q+20=0
q1=2,q2=10/3
∵qd=16-6q
所以:2d=4 d=2 或 10d/3=-4 d=-6/5(舍)
d=2 q=2
an=2n+1 bn=2^n
(2){anbn}为(2n+1)2^n
Tn=(2*1+1)*2^1+(2*2+1)*2^2+.+(2n-1 +1)*2^n-1+(2n+1)2^n
2Tn= (2*1+1)*2^2+(2*2+1)*2^3+.+(2n-2 +1)*2^n-1+(2n-1 +1)*2^n+(2n+1)*2^n+1
-Tn=后面你自己消吧,错位相减应该会吧,太晚了.