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若数列an的前n项和为Sn=2/3an+1/3,则数列an的通项公式是an=?
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若数列an的前n项和为Sn=2/3an+1/3,则数列an的通项公式是an=?
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答案和解析
a(1)=s(1)=(2/3)a(1)+1/3,a(1)=1.
s(n) = (2/3)a(n) + 1/3,
s(n+1)=(2/3)a(n+1)+1/3,
a(n+1) = s(n+1)-s(n)= (2/3)a(n+1) - (2/3)a(n),
a(n+1) = -2a(n),
{a(n)}是首项为1,公比为-2的等比数列.
a(n) = (-2)^(n-1)
s(n) = (2/3)a(n) + 1/3,
s(n+1)=(2/3)a(n+1)+1/3,
a(n+1) = s(n+1)-s(n)= (2/3)a(n+1) - (2/3)a(n),
a(n+1) = -2a(n),
{a(n)}是首项为1,公比为-2的等比数列.
a(n) = (-2)^(n-1)
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