證明：設(shè)三個(gè)連續(xù)奇數(shù)是?$2n - 3，$??$2n - 1，$??$2n + 1(n$?為整數(shù)?$)$?
?$ $?觀察題中的一組式子可得?$(2n - 1)(2n + 1)-(2n - 3)(2n - 1)=4(2n - 1)$?
?$ $?上式左邊?$=(2n - 1)[(2n + 1)-(2n - 3)]$?
?$ =(2n - 1)(2n + 1 - 2n + 3)$?
?$ =(2n - 1)×4$?
?$ =8n - 4$?
?$ $?上式右邊?$=4×(2n - 1)=8n - 4$?
?$ $?左邊?$=$?右邊
?$ $?所以?$(2n - 1)(2n + 1)-(2n - 3)(2n - 1)=4(2n - 1)$?
即上述規(guī)律對(duì)任意三個(gè)連續(xù)奇數(shù)都是成立的
