知 n>1时, 3s(n+1)=a(n+1)+3,①
则 3Sn=an+3 ②
故 ①-② 得 3a(n+1)=a(n+1)-an
即 2a(n+1)=-an a(n+1)/an=-1/2
则数列an是以-1/2为比例的等比数列 且知a1=2
故an=﹛a1=2 (n=1)
an=a1(-1/2)(n-1) =2×(-1/2)(n-1) =-(-1/2)(n-2)
(n-1)、(n-2)均是-1/2的次幂.
知 n>1时, 3s(n+1)=a(n+1)+3,①
则 3Sn=an+3 ②
故 ①-② 得 3a(n+1)=a(n+1)-an
即 2a(n+1)=-an a(n+1)/an=-1/2
则数列an是以-1/2为比例的等比数列 且知a1=2
故an=﹛a1=2 (n=1)
an=a1(-1/2)(n-1) =2×(-1/2)(n-1) =-(-1/2)(n-2)
(n-1)、(n-2)均是-1/2的次幂.