每日一题[855]前n项的积

已知等比数列$\{a_n\}$的首项${a_1} = 1025$,公比$q = - \dfrac{1}{2}$,求${\Pi _n} = {a_1}{a_2} \cdots {a_n}$的最大值.


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分析与解 因为$${a_n} = {a_1} \cdot {q^{n - 1}},$$于是$${\Pi _n} = {a_1}{a_2} \cdots {a_n} = {a_1}^n \cdot {q^{\frac{{n\left( {n - 1} \right)}}{2}}} = {1025^n} \cdot {\left( { - \dfrac{1}{2}} \right)^{\frac{{n\left( {n - 1} \right)}}{2}}}.$$
考虑$$\dfrac{{\left| {{\Pi _{n + 1}}} \right|}}{{\left| {{\Pi _n}} \right|}} = \dfrac{{{{1025}^{n + 1}} \cdot {{\left( {\dfrac{1}{2}} \right)}^{\frac{{n\left( {n + 1} \right)}}{2}}}}}{{{{1025}^n} \cdot {{\left( {\dfrac{1}{2}} \right)}^{\frac{{n\left( {n - 1} \right)}}{2}}}}} = 1025 \cdot {\left( {\dfrac{1}{2}} \right)^n},$$
所以当$n \leqslant 10$时,$$\left| {{\Pi _{n + 1}}} \right| > \left| {{\Pi _n}} \right|,$$当$n \geqslant 11$时,$$\left| {{\Pi _{n + 1}}} \right| < \left| {{\Pi _n}} \right|.$$
当$n$模$4$余$0$或$1$时,${\Pi _n}$$ > 0$;当$n$模$4$余$2$或$3$时,${\Pi _n}$$ < 0$.

所以只需要考虑$n$模$4$余$0$或$1$的情况,因此只需要比较${\Pi _9}$和${\Pi _{12}}$的大小.而$$\dfrac{{{\Pi _{12}}}}{{\Pi { _9}}} = \dfrac{{{{1025}^{12}} \cdot {{\left( {\dfrac{1}{2}} \right)}^{66}}}}{{{{1025}^9} \cdot {{\left( {\dfrac{1}{2}} \right)}^{36}}}} = {1025^3} \cdot {\left( {\dfrac{1}{2}} \right)^{30}} > 1,$$所以$${\Pi _{12}} > {\Pi _9}.$$
因此${\Pi _n}$的最大值为$${\Pi _{12}}= \dfrac{{{{1025}^{12}}}}{{{2^{66}}}}.$$

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