3*(1+2+3+4+.n)-n =3*(1+n)*n/2-n =(3n^2+n)/2
问题描述:
3*(1+2+3+4+.n)-n =3*(1+n)*n/2-n =(3n^2+n)/2
怎么化简到最后一步
答
解由3*(1+2+3+4+.n)-n =3*(1+n)*n/2-n
=(3n+3n^2)/2-n
=(3n+3n^2)/2-2n/2
=(3n+3n^2-2n)/2
=(3n^2+n)/2第二步的-n怎么飞到-2n/2的- -你好这是通分呀你看n=2n÷2=2n/2或-n=(-2n)÷2=-2n/2那最后一步怎么又变加了...(3n+3n^2-2n)/2=(3n-2n+3n^2)/2=[(3-2)n+3n^2]/2=[n+3n^2]/2=(3n^2+n)/2原来如此