在数列an中,a1=1,Sn表示该数列的前n项和,若已知an=2S(n-1),n属于正整数,n大于等于2
问题描述:
在数列an中,a1=1,Sn表示该数列的前n项和,若已知an=2S(n-1),n属于正整数,n大于等于2
求证,数列sn是等比数列.
求数列an的通项公式.
答
an = 2S(n-1)an+S(n-1) = 3S(n-1)Sn=3S(n-1)Sn/S(n-1) =3Sn是等比数列Sn/S(n-1) =3Sn/S1 = 3^(n-1)Sn = 3^(n-1) (1)an =2S(n-1) (2)2(1) -(2)2an = 2.3^(n-1) - anan = (2/3) .3^(n-1)