Question

一道高一数列题,急,已知Sn=1/6(an+1)(an+2) 求an(an>0)

Answer 1

因为Sn=1/6(an+1)(an+2)=1/6(an^2+3an+2)

Sn-1=1/6(an-1^2+3an-1+2)

an=sn-sn-1=1/6(an^2-an-1^2+3an-3an-1)

an^2-an-1^2=3(an+an-1)

(an+an-1)(an-an-1)=3(an+an-1)

因为an>0 an+an-1不等于0

所以 an-an-1=3

公差为3

a1=s1=1/6(a1^2+3a1+2)

(a1-1)(a1-2)=0

所以a1=1或2

an=3n-2或3n-1