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a_n=a_0p^n+\alpha(1-p^n)=a_0p^n+q\sum_{k=0}^{n-1}p^k;\ \ n=1,\ 1-p=0
\sum_{k=1}^{n-1}(loga_{k+1}-loga_k)=\sum_{k=1}^{n-1}log2=log2\sum_{k=1}^{n-1}1=(n-1)log2=loga_n-loga_1
a_{n+1}/a_n=2
a_{n+1}=-a_n\ \ \ \ \ (a_n)=(a_1,a_2,a_3,a_4,..)
a_{n+1}-1=3(a_n -1)=3^{n}(a_1-1)=3^{n+1}(a_0-1)