Question

已知各項(xiàng)均為正數(shù)的數(shù)列{a_n}的前n項(xiàng)和滿足S_n>1,且6S_n=(a_n+1)(a_n+2),n∈N^*.求{a_n}的通項(xiàng)公式.

Answer 1

a_n=3n-1

解:由a₁=S₁=(a₁+1)(a₁+2),
解得a₁=1或a₁=2,由已知a₁=S₁>1,因此a₁=2.
又由a_n+1=S_n+1-S_n=(a_n+1+1)(a_n+1+2)- (a_n+1)(a_n+2),
得(a_n+1+a_n)(a_n+1-a_n-3)=0,
因?yàn)閍_n>0,所以a_n+1-a_n-3=0.
即a_n+1-a_n=3,從而{a_n}是公差為3,首項(xiàng)為2的等差數(shù)列,故{a_n}的通項(xiàng)為a_n=3n-1.