数列{an],n属于N,满足a0=0,a1=2且对任意n属于N有啊a(n+2)=2a(n+1)-an+2,求数列{an}的通项公式2,Tn=1/3a1+1/4a2+……+1/(n+2)an,求Tn
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数列{an],n属于N,满足a0=0,a1=2且对任意n属于N有啊a(n+2)=2a(n+1)-an+2,求数列{an}的通项公式2,Tn=1/3a1+1/4a2+……+1/(n+2)an,求Tn
数列{an],n属于N,满足a0=0,a1=2且对任意n属于N有啊a(n+2)=2a(n+1)-an+2,求数列{an}的通项公式
2,Tn=1/3a1+1/4a2+……+1/(n+2)an,求Tn
数列{an],n属于N,满足a0=0,a1=2且对任意n属于N有啊a(n+2)=2a(n+1)-an+2,求数列{an}的通项公式2,Tn=1/3a1+1/4a2+……+1/(n+2)an,求Tn
a1-a0=2
a(n+2)-a(n+1)=a(n+1)-an+2
∴[a(n+2)-a(n+1)]-[a(n+1)-an]=2
∴数列{a(n+1)-an}是以首项为2,公差为2的等差数列
所以,an-a(n-1)=2+(n-1)×2=2n,.,a2-a1=4,a1-a0=2
把各等式相加可得,an-a0=2+4+6+.+2n=n(2+2n)/2=n(n+1)
即,an=n(n+1)
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