Question

1乘2+2乘3+3乘4+.....+n(n+1)(n+2)(n+3)

急急急急急 快快快快

Answer 1

记an=n(n+1)=n^2+n
则1乘2+2乘3+3乘4+.....+n(n+1)+(n+2)(n+3)是数列｛an｝的前n+1项的和S(n+1)
S(n+1)=1乘2+2乘3+3乘4+.....+n(n+1)+(n+2)(n+3)
=(1^2+1)+(2^2+2)+……+[(n+1)^2+(n+1)]
=[1^2+2^2+……+(n+1)^2]+[1+2+……+(n+1)]
=(n+1)(n+2)(2n+3)/6+(n+1)(n+2)/2
=(n+1)(n+2)(n+3)/3
注： 1^2+2^2+3^2+4^2+5^2+……+n^2=n(n+1)(2n+1)/6
1+2+……+n=n(n+1)]/2

Answer 2

1乘2+2乘3+3乘4+.....+n(n+1)+(n+2)(n+3) 是这个题？

=1*1+1+2*2+2+3*3+3+......n*n+n+（n+1）*（n+1）+n+1
=i²+2²+....+(n+1)²+1+2+....(n+1)
=(n+1)(n+2)(n+3)/6+(n+1)(n+2)/2

追问

恩 帮忙 谢谢