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可看成通項公式an=n×(n+1)的前15項之和,前n項和Sn=1×2+2×3+3×4+4×5+……+n×(n+1)=(1^2+1)+(2^2+2)+(3^2+3)+…+(n^2+n)=(1^2+2^2+3^2+…+n^2)+(1+2+3+…+n)=1/6×n(n+1)(2n+1)+1/2×n(n+1)=1/6…
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