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类型1 a_(n+1)=pa_n+q例1 (2006福建(理))已知数列{a_n}满足a_1=1,a_(n+1)=2a_n+1(n∈N~*),求数列{a_n}的通项公式．解由已知a_(n+1)=2a_(n+1),两边同除以2~(n+1),得(a_(n+1))/(2~(n+1))=(a_n)/(2~n)+1/(2~(n+1))．变形得(a_(n+1))/(2~(n+1))+1/(2~(n+1))=(a_n)/(2~n)+1/(2~n),∴数列{(a_n)/(2~n)+1/(2~n)}是常数列,即(a_n)/(2~n)+1/(2~n)=(a_1)/2+1/2,故所求数列通项为a_n=2~n-1．点拨形如a_(n+1)=pa_n+q(p、q常数,p≠1,q≠0)的递推关系求通项,通常先两边同除以
Type 1 a_(n+1)=pa_n+q Case 1 (Fujian, 2006) Known sequence {a_n} satisfies a_1=1, a_(n+1)=2a_n+1(n∈N~*), Find the general term formula for the series {a_n}. The solution is known by a_(n+1)=2a_(n+1). Dividing both sides by 2~(n+1) gives (a_(n+1))/(2~(n+1))= (a_n)/(2~n)+1/(2~(n+1)). Deformed (a_(n+1))/(2~(n+1))+1/(2~(n+1))=(a_n)/(2~n)+1/(2~n) ∴ The sequence of {(a_n)/(2~n)+1/(2~n)} is a constant column, ie (a_n)/(2~n)+1/(2~n)=(a_1)/2 +1/2, so the general term for the series is a_n=2~n-1. A recurrence relation such as a_(n+1)=pa_n+q(p,q constant,p≠1,q≠0) is used to solve the general term, usually divided by two sides first