已知a为实数,数列{an}满足a1=a,当n≥2时an=an−1−3,(an−1>3)4−an−1,(an−1≤3),
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已知a为实数,数列{an}满足a1=a,当n≥2时a
![已知a为实数,数列{an}满足a1=a,当n≥2时an=an−1−3,(an−1>3)4−an−1,(an−1≤3),](/uploads/image/z/6796315-19-5.jpg?t=%E5%B7%B2%E7%9F%A5a%E4%B8%BA%E5%AE%9E%E6%95%B0%EF%BC%8C%E6%95%B0%E5%88%97%7Ban%7D%E6%BB%A1%E8%B6%B3a1%3Da%EF%BC%8C%E5%BD%93n%E2%89%A52%E6%97%B6an%EF%BC%9Dan%E2%88%921%E2%88%923%EF%BC%8C%28an%E2%88%921%EF%BC%9E3%294%E2%88%92an%E2%88%921%EF%BC%8C%28an%E2%88%921%E2%89%A43%29%EF%BC%8C)
(1)当a=100时,由题意知数列an的前34项成首项为100,公差为-3的等差数列,
从第35项开始,奇数项均为3,偶数项均为1,
从而S100=(100+97+94+…+1)+(3+1+3+1+…+3+1)=
(100+1)×34
2+(3+1)×
66
2=1717+132=1849.
(2)证明:①若0<a1≤3,则题意成立;
②若a1>3,此时数列an的前若干项满足an-an-1=3,即an=a1-3(n-1).
设a1∈(3k,3k+3],(k≥1,k∈N*),
则当n=k+1时,ak+1=a1-3k∈(0,3].
从而,此时命题成立.
综上:对于数列{an},一定存在k∈N*,使0<ak≤3.
从第35项开始,奇数项均为3,偶数项均为1,
从而S100=(100+97+94+…+1)+(3+1+3+1+…+3+1)=
(100+1)×34
2+(3+1)×
66
2=1717+132=1849.
(2)证明:①若0<a1≤3,则题意成立;
②若a1>3,此时数列an的前若干项满足an-an-1=3,即an=a1-3(n-1).
设a1∈(3k,3k+3],(k≥1,k∈N*),
则当n=k+1时,ak+1=a1-3k∈(0,3].
从而,此时命题成立.
综上:对于数列{an},一定存在k∈N*,使0<ak≤3.
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