题目
设等比数列(an)的前n项和为Sn,若 dfrac ({S)_(6)}({S)_(3)}=dfrac (1)(2), 则 dfrac ({S)_(9)}({S)_(3)}= __.
题目解答
答案
解析
步骤 1：设等比数列的首项为a,公比为q,根据题意得: $q\neq 1$
步骤 2：根据等比数列的前n项和公式，有 $S_n = \dfrac{a(1-q^n)}{1-q}$
步骤 3：根据题意，有 $\dfrac{S_6}{S_3} = \dfrac{\dfrac{a(1-q^6)}{1-q}}{\dfrac{a(1-q^3)}{1-q}} = \dfrac{1-q^6}{1-q^3} = \dfrac{1}{2}$
步骤 4：化简上式，得 $1-q^6 = \dfrac{1}{2}(1-q^3)$
步骤 5：解得 $q^3 = -\dfrac{1}{2}$
步骤 6：求 $\dfrac{S_9}{S_3}$，有 $\dfrac{S_9}{S_3} = \dfrac{\dfrac{a(1-q^9)}{1-q}}{\dfrac{a(1-q^3)}{1-q}} = \dfrac{1-q^9}{1-q^3}$
步骤 7：将 $q^3 = -\dfrac{1}{2}$ 代入上式，得 $\dfrac{1-q^9}{1-q^3} = \dfrac{1-(-\dfrac{1}{2})^3}{1-(-\dfrac{1}{2})} = \dfrac{1+\dfrac{1}{8}}{1+\dfrac{1}{2}} = \dfrac{\dfrac{9}{8}}{\dfrac{3}{2}} = \dfrac{3}{4}$
步骤 2：根据等比数列的前n项和公式，有 $S_n = \dfrac{a(1-q^n)}{1-q}$
步骤 3：根据题意，有 $\dfrac{S_6}{S_3} = \dfrac{\dfrac{a(1-q^6)}{1-q}}{\dfrac{a(1-q^3)}{1-q}} = \dfrac{1-q^6}{1-q^3} = \dfrac{1}{2}$
步骤 4：化简上式，得 $1-q^6 = \dfrac{1}{2}(1-q^3)$
步骤 5：解得 $q^3 = -\dfrac{1}{2}$
步骤 6：求 $\dfrac{S_9}{S_3}$，有 $\dfrac{S_9}{S_3} = \dfrac{\dfrac{a(1-q^9)}{1-q}}{\dfrac{a(1-q^3)}{1-q}} = \dfrac{1-q^9}{1-q^3}$
步骤 7：将 $q^3 = -\dfrac{1}{2}$ 代入上式，得 $\dfrac{1-q^9}{1-q^3} = \dfrac{1-(-\dfrac{1}{2})^3}{1-(-\dfrac{1}{2})} = \dfrac{1+\dfrac{1}{8}}{1+\dfrac{1}{2}} = \dfrac{\dfrac{9}{8}}{\dfrac{3}{2}} = \dfrac{3}{4}$