Question

已知数列{a_n}的前n项和为S_n,若S₁=1,S₂=2,且S_n+1-3S_n+2S_n-1=0(n∈N^*且n≥2),求该数列的通项公式.

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),故数列{a_n}从第2项起是以2为公比的等比数列.∴数列{a_n}的通项公式为
a_n=