摘要
例1已知数列{a<sub>n</sub>}中,a<sub>n+1</sub>+a<sub>n</sub>=5,a<sub>1</sub>=2,求数列{a<sub>n</sub>}的通项a<sub>n</sub>.解令a<sub>n</sub>=b<sub>n</sub>+μ,由a<sub>n+1</sub>+a<sub>n</sub>=5,得(b<sub>n+1</sub>+μ)+（b<sub>n</sub>+μ）=5,即b<sub>n+1</sub>=-b<sub>n</sub>+5-2μ,令5-2μ=0,解得μ=5/2.
