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\sum_{k=1}^n(ca_k+db_k)=c\sum_{k=1}^n a_k+d\sum_{k=1}^nb_k
\alpha, \beta
a_{n+1}-\alpha=p(a_n-\alpha)=p^{n+1}(a_0-\alpha)
\alpha=p\alpha+q
a_{n+1}=-a_n+1, a_1=1