通项转换法求和

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

an=n^2+3n
Sn=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

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

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

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

下面我来给你解释一下 平方和公式：1^2+2^2+3^2+.....+n^2=n(n+1)(2n+1）/6
1^2+2^2+3^2+……+n^2=n(n+1)(2n+1)/6

利用立方差公式
n^3-(n-1)^3=1*[n^2+(n-1)^2+n(n-1)]
=n^2+(n-1)^2+n^2-n