Question

19．定義四個(gè)數(shù)a，b，c，d的二階積和式$[\begin{array}{l}ab\\ cd\end{array}]=ad+bc$．九個(gè)數(shù)的三階積和式可用如下方式化為二
階積和式進(jìn)行計(jì)算：$[\begin{array}{l}{a_1}{a_2}{a_3}\\{b_1}{b_2}{b_3}\\{c_1}{c_2}{c_3}\end{array}]={a_1}×[\begin{array}{l}{b_2}{b_3}\\{c_2}{c_3}\end{array}]+{a_2}×[\begin{array}{l}{b_1}{b_3}\\{c_1}{c_3}\end{array}]+{a_3}×[\begin{array}{l}{b_1}{b_2}\\{c_1}{c_2}\end{array}]$．已知函數(shù)f（n）=$[\begin{array}{l}{n}&{2}&{-9}\\{n}&{1}&{n}\\{1}&{2}&{n}\end{array}]$
（n∈N^*），則f（n）的最小值為-21．

Answer 1

分析 根據(jù)定義函數(shù)f（n）=$[\begin{array}{l}{n}&{2}&{-9}\\{n}&{1}&{n}\\{1}&{2}&{n}\end{array}]$=（-9）×$[\begin{array}{l}{n}&{1}\\{1}&{2}\end{array}]$+2×$[\begin{array}{l}{n}&{n}\\{1}&{n}\end{array}]$+n×$[\begin{array}{l}{1}&{n}\\{2}&{n}\end{array}]$=（-9）×（2n+1）+2（n²+n）+n（n+2n）=5n²-16n-9（n∈N^*），根據(jù)二次函數(shù)求出最值．

解答 解：函數(shù)f（n）=$[\begin{array}{l}{n}&{2}&{-9}\\{n}&{1}&{n}\\{1}&{2}&{n}\end{array}]$=（-9）×$[\begin{array}{l}{n}&{1}\\{1}&{2}\end{array}]$+2×$[\begin{array}{l}{n}&{n}\\{1}&{n}\end{array}]$+n×$[\begin{array}{l}{1}&{n}\\{2}&{n}\end{array}]$=（-9）×（2n+1）+2（n²+n）+n（n+2n）=5n²-16n-9
∵n∈N*，∴n=2時(shí)，f（n）的最小值為-21
故答案為：-21

點(diǎn)評(píng) 本題考查了對(duì)新定義的理解，及二次函數(shù)最值問(wèn)題，屬于基礎(chǔ)題，