Question

已知數(shù)列{a_n}的前n項和為S_n,若S₁=1,S₂=2,且S_n+1-3S_n+2S_n-1=0(n∈N^*且n≥2),求該數(shù)列的通項公式.

Answer 1

a_n=

由S₁=1得a₁=1,又由S₂=2可知a₂=1.
∵S_n+1-3S_n+2S_n-1=0(n∈N^*且n≥2),
∴S_n+1-S_n-2S_n+2S_n-1=0(n∈N^*且n≥2),
即(S_n+1-S_n)-2(S_n-S_n-1)=0(n∈N^*且n≥2),
∴a_n+1=2a_n(n∈N^*且n≥2),故數(shù)列{a_n}從第2項起是以2為公比的等比數(shù)列.∴數(shù)列{a_n}的通項公式為
a_n=