已知数列{an}满足a1=2,an=a(n+1)-2,求an及S10
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![已知数列{an}满足a1=2,an=a(n+1)-2,求an及S10](/uploads/image/z/677155-67-5.jpg?t=%E5%B7%B2%E7%9F%A5%E6%95%B0%E5%88%97%7Ban%7D%E6%BB%A1%E8%B6%B3a1%3D2%2Can%3Da%28n%2B1%29-2%2C%E6%B1%82an%E5%8F%8AS10)
已知数列{an}满足a1=2,an=a(n+1)-2,求an及S10
已知数列{an}满足a1=2,an=a(n+1)-2,求an及S10
已知数列{an}满足a1=2,an=a(n+1)-2,求an及S10
a(n+1)-an=2
而a2=a1+2 所以a2=4 a3=6
所以为等差数列 an=2n
sn=n×n+n
s10=110
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