Question

设｛an｝为等差数列,Sn表示它的前n项和,已知对任何正整数n均有Sn=an^2/6+3n/2,求数列｛an｝的首相a1

Answer 1

Sn=an^2/6+3n/2

S1=a1^2/6+3/2=a1

a1^2-6a1+9=0

(a1-3)^2=0

a1=3

s2=a2^2/6+3*2/2=a1+a2

3+a2=a2^2/6+3

a2^2-6a2=0

a2=0,

a2=6

（1）当a2=0时,公差d=a2-a1=0-3=-3

等差数列｛an｝的通项=a1+(n-1)d

an=3-3(n-1)

an=6-3n

（2）当a2=6时,公差d=a2-a1=6-3=3

等差数列｛an｝的通项=a1+(n-1)d

an=3+3(n-1)

an=3n