
1*2=1/3(1*2*3-0*1*2),2*3=1/3(2*3*4-1*2*3),3*4=1/3(3*4*5-2*3*4 求3*[1*2+2*3+3*4+··+n(n+1) 45
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3*[1*2+2*3+3*4+··+n(n+1)]
=3*{1/3*(1*2*3-0*1*2)+1/3*(2*3*4-1*2*3)+1/3*(3*4*5-2*3*4)+........+1/3*[n*(n+1)(n+2)-(n-1)*n*(n+1)]}
=3*1/3*{1*2*3-0*1*2+2*3*4-1*2*3+3*4*5-2*3*4+........+n*(n+1)(n+2)-(n-1)*n*(n+1)}
=3*1/3*{n*(n+1)(n+2)-0*1*2}
=n*(n+1)(n+2)
=3*{1/3*(1*2*3-0*1*2)+1/3*(2*3*4-1*2*3)+1/3*(3*4*5-2*3*4)+........+1/3*[n*(n+1)(n+2)-(n-1)*n*(n+1)]}
=3*1/3*{1*2*3-0*1*2+2*3*4-1*2*3+3*4*5-2*3*4+........+n*(n+1)(n+2)-(n-1)*n*(n+1)}
=3*1/3*{n*(n+1)(n+2)-0*1*2}
=n*(n+1)(n+2)
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