数列an,a1=a2=1,a(n+2)=a(n+1)+tan.(1)t=1时,写出2个能被5整除的项;证明an能被5整除,则a(n+5)也能被5整除.(2)t=2时,求前2n项和S2n
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![数列an,a1=a2=1,a(n+2)=a(n+1)+tan.(1)t=1时,写出2个能被5整除的项;证明an能被5整除,则a(n+5)也能被5整除.(2)t=2时,求前2n项和S2n](/uploads/image/z/4350448-64-8.jpg?t=%E6%95%B0%E5%88%97an%2Ca1%3Da2%3D1%2Ca%EF%BC%88n%2B2%EF%BC%89%3Da%EF%BC%88n%2B1%EF%BC%89%2Btan.%EF%BC%881%EF%BC%89t%3D1%E6%97%B6%2C%E5%86%99%E5%87%BA2%E4%B8%AA%E8%83%BD%E8%A2%AB5%E6%95%B4%E9%99%A4%E7%9A%84%E9%A1%B9%EF%BC%9B%E8%AF%81%E6%98%8Ean%E8%83%BD%E8%A2%AB5%E6%95%B4%E9%99%A4%2C%E5%88%99a%EF%BC%88n%2B5%EF%BC%89%E4%B9%9F%E8%83%BD%E8%A2%AB5%E6%95%B4%E9%99%A4.%EF%BC%882%EF%BC%89t%3D2%E6%97%B6%2C%E6%B1%82%E5%89%8D2n%E9%A1%B9%E5%92%8CS2n)
数列an,a1=a2=1,a(n+2)=a(n+1)+tan.(1)t=1时,写出2个能被5整除的项;证明an能被5整除,则a(n+5)也能被5整除.(2)t=2时,求前2n项和S2n
数列an,a1=a2=1,a(n+2)=a(n+1)+tan.(1)t=1时,写出2个能被5整除的项;证明an能被5整除,则a(n+5
)也能被5整除.
(2)t=2时,求前2n项和S2n
数列an,a1=a2=1,a(n+2)=a(n+1)+tan.(1)t=1时,写出2个能被5整除的项;证明an能被5整除,则a(n+5)也能被5整除.(2)t=2时,求前2n项和S2n
(1)
t=1
a(n+2)=a(n+1)+an
a1=a2=1
a5=5,a10=55能被5整除
若an能被5整除
则a(n+5)=a(n+4)+a(n+3)
=a(n+3)+a(n+2)+a(n+3)
=2a(n+3)+a(n+2)
=2[a(n+2)+a(n+1)]+a(n+2)
=3a(n+2)+2a(n+1)
=3[a(n+1)+an]+2a(n+1)
=5a(n+1)+3an
显然可以被5整除
(2)
t=2
a(n+2)=a(n+1)+2an
则a(n+2)+a(n+1)=2a(n+1)+2an=2[a(n+1)+an]
所以数列{a(n+1)+an}是等比数列,公比是q=2
那么a(n+1)+an=(a2+a1)*q^(n-1)=(1+1)*2^(n-1)=2^n
那么前2n项和S2n=a1+a2+a3+a4+...+a(2n-1)+a(2n)
=(a1+a2)+(a3+a4)+...+[a(2n-1)+a(2n)]
=2^1+2^3+...+2^(2n-1)
=2*(1-4^n)/(1-4)
=2(4^n-1)/3
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