摘要:(Ⅱ)证法一:当n≥2时.因为xn≥>0.xn+1=.
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(2013•茂名一模)已知数列{an},{bn}中,a1=b1=1,且当n≥2时,an-nan-1=0,bn=2bn-1-2n-1.记n的阶乘n(n-1)(n-2)…3•2•1≈n!
(1)求数列{an}的通项公式;
(2)求证:数列{
}为等差数列;
(3)若cn=
+bn-2n,求{cn}的前n项和.
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(1)求数列{an}的通项公式;
(2)求证:数列{
bn |
2n |
(3)若cn=
an |
an+2 |
已知数列{an},{bn}中,a1=b1=1,且当n≥2时,an-nan-1=0,.记n的阶乘n(n-1)(n-2)…3•2•1≈n!
(1)求数列{an}的通项公式;
(2)求证:数列为等差数列;
(3)若,求{cn}的前n项和.
查看习题详情和答案>>
(1)求数列{an}的通项公式;
(2)求证:数列为等差数列;
(3)若,求{cn}的前n项和.
查看习题详情和答案>>
已知数列{an},{bn}中,a1=b1=1,且当n≥2时,an-nan-1=0,bn=2bn-1-2n-1.记n的阶乘n(n-1)(n-2)…3•2•1≈n!
(1)求数列{an}的通项公式;
(2)求证:数列{
}为等差数列;
(3)若cn=
+bn-2n,求{cn}的前n项和.
查看习题详情和答案>>
(1)求数列{an}的通项公式;
(2)求证:数列{
bn |
2n |
(3)若cn=
an |
an+2 |