摘要:12.256.377
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已知{an}为等比数列,a1=1,a5=256;Sn为等差数列{bn}的前n项和,b1=2,5S5=2S8.
(1)求{an}和{bn}的通项公式;
(2)设Tn=a1b1+a2b2+…anbn,求Tn. 查看习题详情和答案>>
(1)求{an}和{bn}的通项公式;
(2)设Tn=a1b1+a2b2+…anbn,求Tn. 查看习题详情和答案>>
设a1,a2,a3,a4,a5为自然数,A={a1,a2,a3,a4,a5},B={a12,a22,a32,a42,a52},且a1<a2<a3<a4<a5,并满足A∩B={a1,a4},a1+a4=10,A∪B中的所有元素之和为256,则集合A为
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{1,2,3,9,12}或{1,3,5,9,11}
{1,2,3,9,12}或{1,3,5,9,11}
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