解
an=n(n+1)=n²+n
∴Sn
=a1+a2+a3+……+an
=1²+1+2²+2+3²+3+……+n²+n
=(1²+2²+3²+……+n²)+(1+2+……+n)
=1/6n(n-1)(2n-1)+1/2(n+1)n
=1/6n[(n-1)(2n-1)+3(n+1)]
=1/6n(2n²-3n+1+3n+3)
=1/6n(2n²+4)
=(n²+2)n/3
公式1²+2²+……+n²=1/6n(n-1)(2n-1)
解
an=n(n+1)=n²+n
∴Sn
=a1+a2+a3+……+an
=1²+1+2²+2+3²+3+……+n²+n
=(1²+2²+3²+……+n²)+(1+2+……+n)
=1/6n(n-1)(2n-1)+1/2(n+1)n
=1/6n[(n-1)(2n-1)+3(n+1)]
=1/6n(2n²-3n+1+3n+3)
=1/6n(2n²+4)
=(n²+2)n/3
公式1²+2²+……+n²=1/6n(n-1)(2n-1)