Question

1×2+2×3+3×4+……n(n+1)

Answer 1

n(n+1)=n^2+n
Sn=1×2+2×3+3×4+……+n(n+1)=1^2+1+2^2+2+3^2+3+……+n^2+n
=(1+2+3+……+n)+(1^2+2^2+3^2+……+n^2)
=n(n+1)/2+(1^2+2^2+3^2+……+n^2)
S（n)=n(n+1)(2n+1)/6
s=1^2+2^2+...+n^2
=n(n+1)(2n+1)/6
=(n^2+n)(2n+1)/6
=(2n^3+3n^2+n)/6

Answer 2

设该数列的通项an=n²+n，则该式子的值为这个数列的前n项的和==
=(1²+2²+……+n²)+(1+2n+……+n)
=n(n+1)(2n+1)/6+n (n+1)/2
剩下的你懂的

Answer 3

1×2=1²+1
2×3=2²+2
3×4=3²+3
……
n(n+1)=n²+n
1×2+2×3+3×4+……n(n+1)
=1²+2²+3²+…+n²+1+2+3+…+n
=n(n+1)(2n+1)/6+(1+n)n/2
=n(n+1)(2n+4)/6
=n(n+1)(n+2)/3
1²+2²+3²+…+n²=n(n+1)(2n+1)/6（一个公式）
1+2+3+…+n=(1+n)n/2（等差数列求和）