Question

数列{n*（n+2）}的前n项和是?

Answer 1

解

n(n+2)=n²+2n

∴sn

=(1²+2²+3²+……+n²)+2(1+2+3+……+n)

=n*(n+1)*(2n+1)/6+2×(1+n)n/2

=1/6n(n+1)(2n+1)+(1+n)n

=n(n+1)(1/3n+1/6+1)

=n(n+1)(1/3n+7/6)

公式1²+2²+……+n²=n*(n+1)*(2n+1)/6