已知数列an中 a1=-2且an+1=sn(n+1为下标),求an,sn

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已知数列an中 a1=-2且an+1=sn(n+1为下标),求an,sn

已知数列an中 a1=-2且an+1=sn(n+1为下标),求an,sn
已知数列an中 a1=-2且an+1=sn(n+1为下标),求an,sn

已知数列an中 a1=-2且an+1=sn(n+1为下标),求an,sn
已知 a_(n+1)=S_n
得 a_n=S_(n-1)(n>1)
两式相减 a_(n+1) - a_n = S_n - S_(n-1) = a_n (n>1)
得 a_(n+1) = 2a_n (n>1)
因为 a_2 = S_1 = a_1 = -2
所以 a_n=-2^(n-1) (n>1,a1=-2);
S_n = a_(n+1) =-2^n

an=sn-1
相减,得到an+1-an=an,也就是an+1=2an
所以an=-2^n
sn=-2^(n+1)