通项转换法求和

显示全部楼层 · 2009-7-25 12:00:28

an=n^2+3nSn=a1+a2+.......+an=(1^2+2^2+......+n^2)+(1*3+2*3+3*3+.......+n*3)=n(n+1)(2n+1)/6 +3*(1+2+3+........+n)=n(n+1)(2n+1)/6 +3(1+n)n/2=(5+n)(n+1)n/3 这里用道平方和公式：1^2+2^2+3^2+.....+n^2=n(n+1)(2n+1）/6

千问 · 2009-7-25 12:00:28

1^2+2^2+3^2+....n^2=n(n+1)(2n+1)/6n*(n+3)=n^2+3n前n项和=1^2+2^2+3^2+....n^2+3*(1+2+3+...+n)=n(n+1)(2n+1)/6+3n(n+1)/2=n(n+1)[(2n+1)/6+3/2]

千问 · 2009-7-25 12:00:28

1^2+2^2+....n^2=n(n+1)(2n+1)/6n*(n+3)=n^2+3nn项和=1^2+2^2+3^2+....n^2+3*(1+2+3+...+n)n(n+1)(2n+1)/6+3n(n+1)/2n(n+1)[(2n+1)/6+3/2] =(5+n)(n+1)n/3

千问 · 2009-7-25 12:00:28

n*(n+3)=n^2+3n∑n*(n+3)=∑（n^2+3n）=∑n^2+3∑n