3. 对于任意有序排列的整式,我们都用右边的整式减去左边的整式,将所得之差的一半写在这两个整式之间,形成一组新的整式,这种操作称为“半路差队”,且把所得到的所有整式之和记为S.现对有序排列的2个整式:
![](http://math.21cnjy.com/mml2svg?mml=%3Cmath+xmlns%3D%22http%3A%2F%2Fwww.w3.org%2F1998%2FMath%2FMathML%22%3E%3Cmrow%3E%3Cmn%3E3%3C%2Fmn%3E%3Cmi%3Ex%3C%2Fmi%3E%3C%2Fmrow%3E%3C%2Fmath%3E)
,
![](http://math.21cnjy.com/mml2svg?mml=%3Cmath+xmlns%3D%22http%3A%2F%2Fwww.w3.org%2F1998%2FMath%2FMathML%22%3E%3Cmrow%3E%3Cmn%3E5%3C%2Fmn%3E%3Cmi%3Ex%3C%2Fmi%3E%3Cmo%3E%2B%3C%2Fmo%3E%3Cmi%3Ey%3C%2Fmi%3E%3C%2Fmrow%3E%3C%2Fmath%3E)
进行“半路差队”操作,可以产生一个新整式串:
![](http://math.21cnjy.com/mml2svg?mml=%3Cmath+xmlns%3D%22http%3A%2F%2Fwww.w3.org%2F1998%2FMath%2FMathML%22%3E%3Cmrow%3E%3Cmn%3E3%3C%2Fmn%3E%3Cmi%3Ex%3C%2Fmi%3E%3C%2Fmrow%3E%3C%2Fmath%3E)
,
![](http://math.21cnjy.com/mml2svg?mml=%3Cmath+xmlns%3D%22http%3A%2F%2Fwww.w3.org%2F1998%2FMath%2FMathML%22%3E%3Cmrow%3E%3Cmi%3Ex%3C%2Fmi%3E%3Cmo%3E%2B%3C%2Fmo%3E%3Cmfrac%3E%3Cmn%3E1%3C%2Fmn%3E%3Cmn%3E2%3C%2Fmn%3E%3C%2Fmfrac%3E%3Cmi%3Ey%3C%2Fmi%3E%3C%2Fmrow%3E%3C%2Fmath%3E)
,
![](http://math.21cnjy.com/mml2svg?mml=%3Cmath+xmlns%3D%22http%3A%2F%2Fwww.w3.org%2F1998%2FMath%2FMathML%22%3E%3Cmrow%3E%3Cmn%3E5%3C%2Fmn%3E%3Cmi%3Ex%3C%2Fmi%3E%3Cmo%3E%2B%3C%2Fmo%3E%3Cmi%3Ey%3C%2Fmi%3E%3C%2Fmrow%3E%3C%2Fmath%3E)
, 记为整式串1,其所有整式之和记为
![](http://math.21cnjy.com/mml2svg?mml=%3Cmath+xmlns%3D%22http%3A%2F%2Fwww.w3.org%2F1998%2FMath%2FMathML%22%3E%3Cmrow%3E%3Cmsub%3E%3Cmi%3ES%3C%2Fmi%3E%3Cmn%3E1%3C%2Fmn%3E%3C%2Fmsub%3E%3C%2Fmrow%3E%3C%2Fmath%3E)
, 则
![](http://math.21cnjy.com/mml2svg?mml=%3Cmath+xmlns%3D%22http%3A%2F%2Fwww.w3.org%2F1998%2FMath%2FMathML%22%3E%3Cmrow%3E%3Cmsub%3E%3Cmi%3ES%3C%2Fmi%3E%3Cmn%3E1%3C%2Fmn%3E%3C%2Fmsub%3E%3Cmo%3E%3D%3C%2Fmo%3E%3Cmn%3E3%3C%2Fmn%3E%3Cmi%3Ex%3C%2Fmi%3E%3Cmo%3E%2B%3C%2Fmo%3E%3Cmfenced+close%3D%22%29%22+open%3D%22%28%22%3E%3Cmrow%3E%3Cmi%3Ex%3C%2Fmi%3E%3Cmo%3E%2B%3C%2Fmo%3E%3Cmfrac%3E%3Cmn%3E1%3C%2Fmn%3E%3Cmn%3E2%3C%2Fmn%3E%3C%2Fmfrac%3E%3Cmi%3Ey%3C%2Fmi%3E%3C%2Fmrow%3E%3C%2Fmfenced%3E%3Cmo%3E%2B%3C%2Fmo%3E%3Cmfenced+close%3D%22%29%22+open%3D%22%28%22%3E%3Cmrow%3E%3Cmn%3E5%3C%2Fmn%3E%3Cmi%3Ex%3C%2Fmi%3E%3Cmo%3E%2B%3C%2Fmo%3E%3Cmi%3Ey%3C%2Fmi%3E%3C%2Fmrow%3E%3C%2Fmfenced%3E%3Cmo%3E%3D%3C%2Fmo%3E%3Cmn%3E9%3C%2Fmn%3E%3Cmi%3Ex%3C%2Fmi%3E%3Cmo%3E%2B%3C%2Fmo%3E%3Cmfrac%3E%3Cmn%3E3%3C%2Fmn%3E%3Cmn%3E2%3C%2Fmn%3E%3C%2Fmfrac%3E%3Cmi%3Ey%3C%2Fmi%3E%3C%2Fmrow%3E%3C%2Fmath%3E)
. 继续对整式串1进行“半路差队”操作,可以得到整式串2,其所有整式之和记为
![](http://math.21cnjy.com/mml2svg?mml=%3Cmath+xmlns%3D%22http%3A%2F%2Fwww.w3.org%2F1998%2FMath%2FMathML%22%3E%3Cmrow%3E%3Cmsub%3E%3Cmi%3ES%3C%2Fmi%3E%3Cmn%3E2%3C%2Fmn%3E%3C%2Fmsub%3E%3C%2Fmrow%3E%3C%2Fmath%3E)
;以此类推,可以得到整式串
![](http://math.21cnjy.com/mml2svg?mml=%3Cmath+xmlns%3D%22http%3A%2F%2Fwww.w3.org%2F1998%2FMath%2FMathML%22%3E%3Cmi%3En%3C%2Fmi%3E%3C%2Fmath%3E)
, 其所有整式之和记为
![](http://math.21cnjy.com/mml2svg?mml=%3Cmath+xmlns%3D%22http%3A%2F%2Fwww.w3.org%2F1998%2FMath%2FMathML%22%3E%3Cmrow%3E%3Cmsub%3E%3Cmi%3ES%3C%2Fmi%3E%3Cmi%3En%3C%2Fmi%3E%3C%2Fmsub%3E%3C%2Fmrow%3E%3C%2Fmath%3E)
. 下列说法:
①整式串4共有18个整式;
②第2022次操作后,所有整式之和为
;
③若
, 则
.
其中正确的个数是( )