等差数列an各项均为正数,a1=3前n项和为sn,等比数列bn中,b1=1且b2s2=64 {ban}是公比为64的等比
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等差数列an各项均为正数,a1=3前n项和为sn,等比数列bn中,b1=1且b2s2=64 {ban}是公比为64的等比数列.
(1)求an与bn (2)求{anbn}的前n项和Tn
(1)求an与bn (2)求{anbn}的前n项和Tn
![等差数列an各项均为正数,a1=3前n项和为sn,等比数列bn中,b1=1且b2s2=64 {ban}是公比为64的等比](/uploads/image/z/1756652-68-2.jpg?t=%E7%AD%89%E5%B7%AE%E6%95%B0%E5%88%97an%E5%90%84%E9%A1%B9%E5%9D%87%E4%B8%BA%E6%AD%A3%E6%95%B0%2Ca1%3D3%E5%89%8Dn%E9%A1%B9%E5%92%8C%E4%B8%BAsn%2C%E7%AD%89%E6%AF%94%E6%95%B0%E5%88%97bn%E4%B8%AD%2Cb1%3D1%E4%B8%94b2s2%3D64+%7Bban%7D%E6%98%AF%E5%85%AC%E6%AF%94%E4%B8%BA64%E7%9A%84%E7%AD%89%E6%AF%94)
设an的公差为d,bn的公比为q
a2=a1+d=3+d,b2=b1*q=q
ban/ba(n-1)=q^(an-a(n-1))=q^d=64(明显q不等于1)
b2s2=64
6q+dq=64,
an各项均为正数,d>0且为正整数,所以64/q为正整数且大于6,q可能取值为2,4,8
带入d=64/q-6,d对应取值为26,10,2
符合q^d=64的只有q=8,d=2
所以an=3+2(n-1)=2n+1,bn=8^(n-1)
Tn=a1b1+a2b2+.+anbn 等式(1)两边都乘以公比q
得到qTn=a1b2+a2b3+.+等式(2)
(1)-(2)
-7Tn=a1b1+(a2-a1)b2+(a3-a2)b3+.+(an-a(n-1))bn-
=a1b1+b2+b3+.+bn-anb(n+1)=2+(b1+b2+...+bn)-anb(n+1)
=2+(8^n-1)/7-(2n+1)*8^n
Tn=(2n+1)*8^n/7-(8^n-1)/49-2/49
a2=a1+d=3+d,b2=b1*q=q
ban/ba(n-1)=q^(an-a(n-1))=q^d=64(明显q不等于1)
b2s2=64
6q+dq=64,
an各项均为正数,d>0且为正整数,所以64/q为正整数且大于6,q可能取值为2,4,8
带入d=64/q-6,d对应取值为26,10,2
符合q^d=64的只有q=8,d=2
所以an=3+2(n-1)=2n+1,bn=8^(n-1)
Tn=a1b1+a2b2+.+anbn 等式(1)两边都乘以公比q
得到qTn=a1b2+a2b3+.+等式(2)
(1)-(2)
-7Tn=a1b1+(a2-a1)b2+(a3-a2)b3+.+(an-a(n-1))bn-
=a1b1+b2+b3+.+bn-anb(n+1)=2+(b1+b2+...+bn)-anb(n+1)
=2+(8^n-1)/7-(2n+1)*8^n
Tn=(2n+1)*8^n/7-(8^n-1)/49-2/49
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