若数列｛An}的前n项和Ｓn=2An+1,问数列｛Ａn｝是否成等比数列？说明理由，并求出An
热心网友
Ｓn=2An+1 == Ｓn-1 =2A(n-1)+1 == Ｓn - Ｓn-1 = An = (2An+1)-[2A(n-1)+1] = 2An -2A(n-1)An/A(n-1) = 2== ｛Ａn｝成等比数列Ｓ1 = A1 = 2A1 +1 == A1 = -1== An = 1 - 2^n
若数列｛An}的前n项和Ｓn=2An+1,问数列｛Ａn｝是否成等比数列？说明理由，并求出An
Ｓn=2An+1 == Ｓn-1 =2A(n-1)+1 == Ｓn - Ｓn-1 = An = (2An+1)-[2A(n-1)+1] = 2An -2A(n-1)An/A(n-1) = 2== ｛Ａn｝成等比数列Ｓ1 = A1 = 2A1 +1 == A1 = -1== An = 1 - 2^n