1×2+2×3+3×4+4×5+…n(n+1)=1×(1+1)+2×(2+1)+3×(3+1)+4×(4+1)+…n(n+1)=1²+1+2²+2+3²+3+4²+4+…+n²+n=1²+2²+3²+4²+…+n²+1+2+3+4+…n=(1/6)n(n+1)(2n+1)+(1/2)n(n+1)=(1/6)n(n+1)(2n+1+3)=(1/3)n(n+1)(n+2)
