已知数列{an}满足a1=2,an=an+1再-2,求an和S10
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已知数列{an}满足a1=2,an=an+1再-2,求an和S10
已知数列{an}满足a1=2,an=an+1再-2,求an和S10
已知数列{an}满足a1=2,an=an+1再-2,求an和S10
A(n+1)-An=2
∴An是首项为2,公差为2的等差数列
an=2+2(n-)=2n
Sn=na1+1/2n(n-1)d=2n+1/2n(n-1)*2=n²+n
s10=10²+10=110
an=an+1 -2 所以是一个等差数列 所以通项an=2+2(n-1) S10=10a1+(1+2....+9)d=10*2+45*2=110
1+2+...9=n(n+1)/2 =9*10/2=45
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